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THE NEW 

METAL WORKER 
PATTERN B 

A Complete Course of Instruction in the Modern Methods of Developing 
and Cutting the Patterns for Sheet Metal Work, Giving the Prin- 
ciples Underlying Practically every Problem that is Likely to 
Come up in Practice and Explaining the Selection and Use of 
Drawing Tools and Linear and Geometrical Drawing so 
Clearly that One who has had no Previous Knowledge 
of Arithmetic or Drawing, may Understand 
These Essentials and Apply Them in Using 
the 253 Problems in the Development 
of Patterns by the Parallel Line, 
Conical or Flaring, and 
Irregular orTriangula- 
tion Systems. 



BY 

GEO. W? KITTREDGE and ASSOCIATES 






New York 

U. P. C. BOOK COMPANY, Inc. 

241-249 West 39th Street 

Nineteen Hundred Seventeen 



5* S ° 



Copyrighted, 1917 

By 

U. P. C. Book Company, Inc. 

Copyrighted, 1896, 1906 

By 

David Williams Company 



) 7-^<? } 



• : 






OCT 31 1917 



PRESS OF 

BRAUNWORTH & CO. 

BOOK MANUFACTURERS 

BROOKLYN, N. Y. 



ICI.A477338 



PREFACE TO 1917 EDITION. 



That the New Metal Worker Pattern Book has rilled and still continues to fill a real need in 
modern sheet metal shops is fully attested by its long and tremendous sale. Ever since the pub- 
lication of the first edition it has been considered the standard authority on sheet metal pattern 
cutting, and many affectionately term it "The Bible of the Trade." 

No better textbook for home study is available, and, even though the student does not know a 
thing about arithmetic or drawing, he may soon master both linear and geometrical drawing if he will 
diligently study the chapters on these subjects. The Secret of Success in sheet metal pattern drafting is in 
knowing how to apply the principles of geometry to your problem. If you will drill yourself carefully in 
both linear and geometrical drawing, you will never be stuck, no matter how difficult a job you may get. 
The facts given in these two chapters of the book may help you to make good. Therefore, study them 
now and keep at it until you know them as you do your tools. 

The pattern problems are arranged so as to take the student from the most elementary work, by 
easy stages, through every branch of sheet metal work to the development of the most difficult problems 
by the modern system of Triangulation. This is the only book on the market at present giving a simple 
and practical explanation of the use of Triangulation. Sixty-five problems comprising figures and over 250 
pages are devoted to the subject, without counting those in which Triangulation and either the Parallel 
line or Conical forms are combined. 

In the former edition the problems were classified according to the three forms or methods used in 
their development — that is, Parallel line, Conical, flaring or radial fine development, and the modern sys- 
tem of Triangulation. For the past few years, however, experts have been working on short-cut methods, 
and as many of these require the use of more than one of the three forms, it has been found necessary 
in arranging these new problems to add a fourth section on Mixed or Combination Forms to this new 
edition. 

These new methods and many examples of the application of Triangulation and other forms of 
development in such modern work as automobiles, marquees, grain chutes, dust separators, conveyors, 
duct work, furnace fittings, flanges for flag poles, boxes, blower connections, range canopies, hoppers, spouts, 
etc., form a section which is well worth the price of the complete book. 

These four pattern problem sections form the most complete reference library for the busy pattern 
draftsman or cutter that has ever been published. No matter how difficult his problem he is sure to 
find help and inspiration to its solution. 

Those who would like additional examples of the application of the principles demonstrated in this 
work will find the series of twelve volumes entitled "Practical Sheet Metal Work and Demonstrated 
Patterns" well worth having, for each volume in that series covers a different branch of the trade, such as 
leaders, gutters, roofing, cornice work, skylights, automobiles, etc. and gives not only pattern problems 
but a great fund of material on both shop and outside practice. Not one of the problems are duplicates 
of those given in this book. 

This new, greatly enlarged edition of the Pattern Book has been carefully corrected and it is offered 
to the trade with the hope and expectation that it will prove an infallible aid to everyone interested in 
sheet metal pattern drafting. 



PREFACE. 



FOR the benefit of those who may contemplate making use of this work, wholly or in part, it is 
well to lay before them at the outset a general statement of the plan upon which it is written, 
together with some advice for the use and study of the same, which may not properly belong 
under any of the several headings comprising the subject matter. A glance at the table of con- 
tents immediately following will give at once a clear idea of its scope and arrangement. From 
this it will be seen that the first five chapters are theoretical or educational in their nature, 
while the last chapter is devoted to practical work ; and further, that the book does not presume 
upon any previous technical knowledge upon the part of the beginner, but aims to place before 
him in the preliminary chapters all that is necessary to a thorough understanding of the work 
performed in the last chapter, which constitutes the bulk of the book. 

A very important feature of the work is the classification of the problems. The forms for 
which patterns may be required are divided, according to the methods employed in developing 
their surfaces, into three classes, and the problems relating to each are arranged in three corre- 
sponding sections of the last chapter, thus bringing near together those in which principles and 
methods are alike. In Chapter V (Principles of Pattern Cutting) this classification is defined and 
the principles governing each class are explained and illustrated under three sub-headings of the 
chapter. The third subdivision treats of the method of developing the surfaces of irregular 
forms by Triangulation, a subject not heretofore systematically treated in any work on pattern 
cutting. 

A chapter on drawing (Chapter III) has been prepared for the benefit of the pattern cutter 
especially interested in cornice work, and though he may not intend to become a finished archi- 
tectural draftsman, this chapter will render him valuable assistance in reading the original drawings 
received from^ architects, from which he is required in many cases to make new drawings adapted 
to his own peculiar wants. 

The New Metal Worker Pattern Book, besides being a systematic treatise on the principles of 
pattern cutting, is also valuable as a reference book of pattern problems and as a fund of information 
on the subject treated, to be drawn from at convenience, and is so written that each problem, or chapter 
of descriptive matter, can be read independently of the others; so that the student whose time is 
limited can turn to any portion of the work the title of which promises the information sought, 
without feeling that he must read all that precedes it. The relative importance of the chapters 
depends, of course, upon the individual reader, and will be determined by what he considers his 
weakest points. However, it is advisable in the study of all works of a scientific nature to begin 
at the beginning and take everything in its course. If, therefore, the study of this work can be 
continued progressively from the first, much advantage will be gained. 

The statement of each problem in prominent type appears at the head of the demonstration, 



The New Metal Worker Pattern Book. 

and every problem is numbered, by which arrangement the problems are well separated from each 
other and easily found. 

While each demonstration is considered complete in itself, some are necessarily carried farther 
into detail than others, and references are made from one problem to another, pointing out similarity 
of principle, where such comparison would be advantageous to one who is looking for principles 
rather than for individual solutions. 

In preparing the diagrams used to illustrate the solutions of the problems, forms have been 
chosen which are as simple in outline as the case will admit, upon the supposition that the reader 
will be able to make the application of the method described in connection with the same to his own 
especial case, which may embody more complicated forms. It must also be noted that, owing to the 
small scale to which the drawings in this work are necessarily made, extreme accuracy in the 
operations there performed is impossible. In many instances the length of the spaces used in 
dividing the profiles is much too great in proportion to the amount of curvature to insure accuracy. 
Therefore if apparent errors in measurements or results are found, they must not be considered the 
fault of the system taught. If such errors are discovered the student is recommended to reconstruct 
the drawing upon his own drawing board in accordance with the demonstration given and to a scale 
sufficiently large to insure accurate results, before passing judgment. 

In the preparation of this book, the former Metal Worker Pattern Book has been made the basis, 
to a certain extent, of the new work.* Such problems or portions of the former work as were 
found satisfactory have been assigned to their proper places in the new work without change. In 
the case of most of the problems, however, the demonstrations have been revised and the drawings 
accompanying them have been amended or corrected in accordance with the text, and in many cases 
entire new drawings have been made. To these have been added a large number of new problems 
based upon inquiries and solutions that have appeared in the columns of The Metal Worker since the 
former work was published. Much new explanatory matter not in the former work has also been 
added in the preliminary chapters, prominent among which are Chapter III, and the principles of 
Triangulation in Chapter V. 

Especial care has been taken in the composition of the book to have each engraving and the 
text referring to it arranged, as far as possible, on the same page or upon facing pages, so as to 
obviate the necessity of turning the leaf in making references. 

A great advantage is gained over the former work by the classification and numbering of the 
problems, which, in connection with the table of contents, renders any desired subject or problem 
easily found. 

In regard to the system of reference letters emjuoyed in the drawings, it should be said that the 
same letter has been used so far as possible to represent any given point in the several views or 
positions in which it may occur, the superior figure or exponent being changed in each view. To 
fully comprehend this the reader must carry in mind the concrete idea of the form under considera- 
tion, just as though he held in his hand a perfect completed model of the same, which he turned 
this way or that to obtain the several views given. Any point, therefore, which might on the model 
be marked by a letter A, would be designated in one of the views as A, while in other views or 
places where it might appear it would be designated as A 1 , A 2 , etc., or as A', A", etc. In the 

* Publisher's Note. — The author of this book, George W. Kittredge, prepared the drawings and outlined the 
demonstrations of all but a few of the less important problems in The Metal Worker Pattern Book, which was 
published in 1881, and also prepared portions of the introductory chapters of that work. 



vi Introduction. 

solution of problems by triangulation, dotted lines are alternated with solid lines, as lines of meas- 
urement, merely for the sake of distinction and to facilitate the work. 

Occasions arise in the experience of every pattern cutter wherein some portion of the work 
before him, of relatively small importance, is so situated that the development of its pattern by a 
strictly accurate method would involve more labor and time than would be justified by the value 
of the part wanted. It is the purpose of this work to teach the principles of pattern cutting, leav- 
ing the decision of such questions to the individual. Nevertheless, if one is thoroughly con- 
versant with pattern cutting methods and familiar with pattern shapes it may be possible in such 
cases to obtain accurately the principal points of a required pattern and to complete the same by the 
eye with sufficient accuracy for all practical purposes. 

As intimated above, some of the demonstrations are necessarily made more explicit than others. 
In the longer demonstrations and those occurring near the ends of the Sections, less important details 
of the work are sometimes omitted and certain parts of the operation are only hinted at or are 
described in a general way, upon the supposition that the simpler problems in which the demon- 
strations are carried further into detail would naturally be studied first. 

Although the principles of pattern cutting here set forth may at times be regarded as somewhat 
intricate, it is believed that any one possessed of a fair degree of intelligence and application can 
easily master them. 

Notwithstanding the great care which has been used in the preparation of this work, it is 
possible that errors may have found their way into its columns. Should errors be discovered by any 
of its readers, information of such will be gladly received. 



CONTENTS. 



CHAPTER I. 

a*AGB 

Terms and Definitions „ „ ........... 1 

Alphabetical List of Terms 15 

CHAPTER II. 

Drawing Instruments and Materials 17 

CHAPTER III. 

Linear Drawing 30 

CHAPTER IV. 

Geometrical Problems 36 

Construction of Regular Polygons 43 

The Ellipse . . 59 

The Volute .... 67 

CHAPTER V. 

Principles of Pattern Cutting 71 

Parallel Forms 72 

Regular Tapering Forms 79 

Irregular Forms . 86 

CHAPTER VI. 

Pattern Problems 96 

Section 1. — Parallel Forms (Miter Cutting) 96 

Section 2. — Regular Tapering Forms (Flaring Work) 240 

Section 3. — Irregular Forms (Triangulation) 306 

Section 4. — Mixed or Combination Forms 448 

Index of Problems ._ „5ig 



CHAPTER I. 



Terms and DefleitioeSc 



Pattern cutting as applied to sheet-metal work, 
by its very nature, involves the application of geo- 
metrical principles. Any treatise on descriptive geom- 
etry presents in a general way all the principles that 
enter into the science of pattern cutting. To those who 
have had the advantages of a mathematical education 
these principles are well known and by such their ap- 
plication is easily made. For the benefit of those, 
however, who have not had such advantages, this work 
purposes to make specific application of those princi- 
ples in a way to be readily understood by the mechanic. 
While throughout the work the use of an unnecessary 
number of technical terms and words not in common 
use among mechanics will be carefully avoided, it must 
be here noted that precise language in describing all 
geometrical figures and operations becomes a necessity, 
and therefore compels the employment of some terms 
not in the every day vocabulary of the workshop, which 
it is proper to define and explain at the outset. As 
the language of the workshop is usually far from ac- 
curate and varies with the locality, every student of 
this book will find it greatly to his advantage to give 
careful attention to this and the other introductory 
chapters for the purpose of increasing and improving 
his vocabulary, and of enabling him to more readily 
comprehend the demonstrations in the pages following. 
The list of terms herein defined has not been restricted 
to the barest requirements of the book, but has been 
made to include nearly all the terms belonging to plane 
geometry, and such architectural terms as are usually 
met with in problems relating to cornice work. The 
terms are arranged first logically, in classes, after which 
follows an alphabetical list by which any definition can 
be readily found. 

1. Geometry is that branch of mathematics which 
treats of the relations, properties and measurements of 
lines, angles, surfaces and solids. 

2. Sheet-Metal Pattern Cutting: is founded upon 
those principles of geometry which relate to the sur- 



faces of solids, and may be more accurately described 
as the development of surfaces, under which name its 
principles are now being taught to a great extent in 
schools of practical instruction. Articles made from 
sheet metal are hollow, being only shells, and must, 
therefore, be considered in the process of pattern cut- 
ting as though they were the coverings or casings 
stripped from solids of the same shape. 

3. A Point is that which has place or position with- 
out magnitude, as the intersection of two lines or the 
center of a circle ; it is usually represented to the eye 
by a small dot. 

LINES. 

4. A Line is that which has length merely, and 
may be straight or curved. 

5. A Straight Line, or, as it is sometimes called, a 
right line, is the shortest line that can be drawn between 
two given points. Straight lines are generally desig- 
nated by letters or figures at their extremities, as A B, 
Fig. 1. 

6. A Curved Line is one which changes its direc- 
tion at every point, or one of which no portion, how- 

A B 



Fig. 1 —A Straight Line. 

ever small, is straight. It is therefore longer than a 
straight line connecting the same points. Curved lines 




Fig. 2. — Curved Lines. 

are designated by letters or figures at their extrem- 
ities and at intermediate points, as A B C or D E F, 
Fig. 2. 



The New Metal Worker Pattern Book. 



7. Parallel Lines are those which have no inclina- 
tion to each other, being everywhere equidistant. A 
B and A' B 1 in Fig. 3 are parallel straight lines, and 



- B 

i 

-a 




C 1 D 

Fig. 3. — Parallel Lines. 

can never meet though produced to infinity. C D and 
C D 1 are parallel curved lines, being arcs of circles 
which have a common center. 

8. Horizontal Lines are lines parallel to the hori- 
zon, or level. A Horizontal Line in a drawing is indi- 
cated by a line drawn from left to right across the 



paper. 



as A B in Fig. 4. 




Fig. If. — Names of Lines by Direction. 

9. Vertical Lines are lines parallel to a plumb 
line suspended freely in a still atmosphere. A Verti- 
cal Line in a drawing is represented by a line drawn 
up and down the paper, or at right angles to a hori- 
zontal line, as E C in Fig. 4. 

10. Inclined or Oblique Lines occupy an interme- 
diate between horizontal and vertical lines, as D, 



Fig. 5. — Perpendicular Lines. 

Fig. 4. Two lines which converge toward each other 
and which, if produced, would meet or intersect, are 
said to incline to each other. 



11. Perpendicular Lines — Lines are perpendicular 
to each other when the angles on either side of the 
point of meeting are equal. Vertical and horizontal 
lines are always perpendicular to each other, but per- 
pendicular lines are not always vertical and horizontal, 
but may be at any inclination to the horizon, provided 
that the angles on either side of the point of intersec- 
tion are equal. In Fig. 5, C F, D H and E Gr are 
said to be perpendicular to A B. Also in Fig. 6, 




B ■ F 

Fig. 6. — Perpendicular Lines. 

C D and E F are perpendicular to A B. Lines per- 
pendicular to the same line are parallel to each other, 
as C F and D H, Fig. 5, which are perpendicular 
to A B. 

12. An Angle is the opening between two straight 
lines which meet one another. An angle is commonly 
designated by three letters, the letter designating the 
point in which the straight lines containing the angle 




Fig. 



-Angles. 



meet being between the other two letters, as the angle 
ECD, Fig. 4. 

13. A Right Angle.— When a straight line meets 
another straight line so as to make the adjacent angles 
equal to each other, each angle is a right angle, and 
the straight lines are said to be perpendicular to each 
other. (See CBEorCBD, Fig. 7.) 

14. An Acute Angle is an angle less than a right 
angle, as A B D or A B C, Fig. 7. 

15. An Obtuse Angle is an angle greater than a 
right angle, as A B E, Fig. 7. 



Terms and Definitions. 



3 



STRAIGHT SIDED FIGURES. 

16. A Surface is that winch has length and 
breadth without thickness. 

17. A Plane is a surface such that if any two of 
its points be joined by a straight line, such line will be 
wholly in the surface. Every surface which is not a 
plane surface, or composed of plane surfaces, is a 
curved surface. 

18. A Single Curved Surface is one in which only 
certain points may be joined by straight lines which 





Fig. 8. — An Equilateral Triangle. Fig. 9. — An Isosceles Triangle 
B D 




Fig. 10.— A Scalene Triangle. Fig 11.— Right-Angled Triangles. 

shall lie wholly in its surface. The rounded surface of 
a cylinder or cone is a single curved surface. 

19. A Double Curved Surface is one in which no 
two points can be joined by a straight line lying wholly 
in its surface. The surface of a sphere, for example, 
is a double curved surface. 

20. A Plane Figure is a portion of a plane termi- 
nated on all sides by lines either straight or curved. 

21. A Rectilinear Figure is a surface bounded by 
straight lines. (See Figs. 8, 16, 21, etc.) 

22. Polygon is the general name applied to all 
rectilinear figures, but is commonly applied to those 
having more than four sides. A regular polygon is one 
in which the sides are equal. 

23. A Triangle is a flat surface bounded by three 
straight lines. (Figs. 8, 9, 10, 11, 13, etc.) 

24. An Equilateral Triangle is one in which the 
three sides are equal. (Fig. 8.) 



25. An Isosceles Triangle is one in which two of 
the sides are equal. (Fig. 9.) 

26. A Scalene Triangle is one in which the three 
sides are of different lengths. (Fig. 10.) 

27. A Right-Angled Triangle is one in which one 
of the angles is a right angle. (Fig. 11.) 

28. An Acute-Angled Triangle is one which has its 
three angles acute. (Fig. 12.) 

29. An Obtuse-Angled Triangle is one which has 
an obtuse angle. (Fig 13.) 




Fig. IS.— An Acute-Angled 
Triangle. 



Fig. IS. — An Obtuse-Angled 
Triangle. 




Apex orVerteX 
By 




Fig. U.— Names of the Sides of a Fig. 15.— Names of the Parts 
Right-Angled Triangle. of a Triangle. 

30. A Hypothenuse is the longest side in a right- 
angled triangle, or the side opposite the right an°-le. 
A C, Fig. 14. 

31. The Apex of a triangle is its upper extremity, 
as B, Fig. 15. It is also called vertex. 

32. The Base of a triangle is the line at the bottom. 
B C and A C, Figs. 14 and 15. 

33. The Sides of a triangle are the including lines. 
A C, A B and B C, Figs. 14 and 15. 

34. The Vertex is the point in any figure opposite 
to and furthest from the base. The vertex of an an<de 
is the point in which the sides of the angle meet. B 
Fig. 15. 

35. The Altitude of a triangle is the length of a 
perpendicular let fall from its vertex to its base, as B 
D, Fig. 15, 

36. A Quadrilateral figure is a surface bounded by 
four straight lines. There are three kinds of Quadri- 



TJie New Metal Worker Pattern Book. 



laterals : The Trapezium, the Trapezoid and the Par- 
allelogram. 

37. The Trapezium has no two of its sides parallel. 
(Fig. 16.) 

38. The Trapezoid has only two of its sides parallel. 
(Fig. IT.) 

39. The Parallelogram has its opposite sides par- 
allel. There are four varieties of parallelograms : The 
Rhomboid, the Rhombus, the Rectangle and the 
Square. 



46. A Heptagon is a plane figure of seven sides. 
(Fig. 24.) 

47. An Octagon is a plane figure of eight sides. 
(Fig. 25.) 

48. A Decagon is a plane figure of ten sides. 
(Fig. 26.) 

49. A Dodecagon is a plane figure of twelve sides. 
(Fig. 27.) 

50. The Perimeter is the line or lines bounding any 
figure, as A B C D E, Fig. 22. 





Fig. 16.— A Trapezium. 



Fig. 17.— A Trapezoid. 





Fig. 22.— A Pentagon. 

wmm. 



Fig. 2S.—A Hexagon. Fig. 21,.— A Heptagon. 






Fig. 18.— A Rhomboid. 



Fig. 19. — A Rhombus or Lozenge. 



Fig. 25.— An Octagon. 



Fig. 26.— A Decagon. Fig. 27.— A Dodecagon. 





Fig. 20.— An Equiangular Par- Fig. 21. — An Equiangular 
allelogram Called a Rectangle. and Equilateral Parallel- A 

ogram Called a Square. Fig. 28. — Diagonals. 





Fig. 29.— A Circle. 



40. The Rhomboid has only the opposite sides 
equal, the length and width being different and its 
angles are not right angles. (Fig. 18.) 

41. The Bhombus, Lozenge or diamond is a rhom- 
boid all of whose sides are equal. (Fig. 19.) 

42. The Rectangle is a parallelogram all of whose 
angles are right angles. (Fig. 20.) 

43. The Square is an equilateral rectangle. (Fig. 
21.) 

44. A Pentagon is a plane figure of five sides. 
(Fig. 22.) 

45. A Hexagon is a plane figure of six sides 
(Fig. 23.) 



51. A Diagonal is a straight line joining two oppo- 
site angles of a figure, as A B and C D, Fig. 28. 

CIRCLES AND THEIR PROPERTIES. 

52. A Circle is a plane figure bounded by a curved 
line, everywhere equidistant from its center. (Fig. 29.) 
The term circle is also used to designate the boundary 
line. (See also Circumference ) 

53. The Circumference of a circle is the boundary 
line of the figure. (Fig. 29.) 

54. The Center of a circle is a point within the 
circumference equally distant from every point in its 

I circumference, as A, Fig. 29. 



Terms and Definitions. 



55. The Radius of a circle is a line drawn from 
the center to any point in the circumference, as A B, 
Fig. 29, that is, half the diameter. The plural of radius 
is radii. 

56. The Diameter of a circle is any straight line 
drawn through the center to opposite points of the cir- 
cumference, as C D, Fig. 29. 

57. A Semicircle is the half of a circle, and is 




Diameter 
Fig SO. — A Semicircle 



Fig. SI — Segments. 



bounded by half the circumference and a diameter. 
(Fig. 30.) * 

58. A Segment of a circle is any part of its sur- 
face cut off by a straight line, as A E B and CFD, 
Fig. 31. 

59. An Arc of a circle is any part of the circum- 
ference, as A B E and C F D, Fig. 32. 





Fig. SS. — Arcs and Chords. 



Fig. 33.— Sectors. 



60. A Chord is a straight line joining the extrem- 
ities of an arc, as A E and C D, Fig. 32. 

61. A Sector of a circle is the space included be- 
tween two radii and the arc which they intercept, as 
A C B and DCE, Fig. 33, and B A C, Fig. 34. 

62. A Quadrant is a sector whose area is equal to 
one-fourth of the circle. (B A C, Fig. 34.) The two 
radii bounding a quadrant are at right angles. 

63. A Tangent to a circle or other curve is a. 
straight line which touches it at only one point, as E D 
and A C, Fig. 35. Every tangent to a circle is perpen- 



dicular to the radius drawn to the point of tan- 
gency. Thus E D is perpendicular to F D and A G 
toFB. 




Fig 34.— A Quadrant. 



Fig. 35.— Tangents. 



64. Concentric circles are those which are described 
about the same center. (Fig. 36.) 

65. Eccentric circles are those which are described 
about different centers. (Fig. 37.) 

66. Polygons are inscribed in, or circumscribed by, 





Fig. 36. Concentric Circles. 



Fig 37. — Eccentric Circles. 



circles when the vertices of all their angles are in the 
circumference. (Fig. 38.) 

67. A circle is inscribed in a straight-sided figure 
when it is tangent to all sides. (Fig. 39.) All regular 
polygons may be inscribed in circles, and circles may 





X A d 

Fig. 38.— An Inscribed Triangle. Fig. S9.—An Inscribed Circle. 

be inscribed in the polygons ; hence the facility with 
which polygons may be constructed. 

68. A Degree. — The circumference of a circle is 
considered as divided into 360 equal parts, called degrees 



6 



The New Metal Worker Pattern Book. 



(marked °). Each degree is divided into 60 minutes 
(marked ') ; and each minute into 60 seconds (marked "). 
Thus if the circle be large or small the number of 
divisions is always the same, a degree being equal to 




Fig. 40 —A Circle Divided into Degrees for Measuring Angles. 

•j^th part of the whole circumference ; the semicircle 
is equal to 180° and the quadrant to 90°. The radii 
drawn from the center of a circle to the extremities of 
a quadrant are always at right angles with each other ; 
a right angle is therefore called an angle of 90° (A E 
B, Fig. 40). If a right angle be bisected by a straight 
line, it divides the arc of the quadrant also into two 
equal parts, each being equal to one-eighth of the 
whole circumference, or 45°, (A E F and FEB, Fig. 
40) ; if the right angle were divided into three equal 
parts by straight lines, it would divide the arc into 
three equal parts, each containing 30° (A EG, G E H, 
H E B, Fig. 40). Thus the degrees of the circle are 





v .y 

Fig. 41.— Complement. 



Fig. 4%. — Supplement. 



used to measure angles, therefore by an angle of any 
number of degrees, it is understood that if a circle with 
any length of radius be struck with one foot of the 
compasses in its vertex, the sides of the angle will 



intercept a portion of the circle equal to the num. 
ber of degrees given. Thus the angle A E H, Fig. 
40, is an angle of 60°. In the measurement of angles 
by the circumference of the circle, and in the various 
mathematical calculations based thereon, use is made 
of certain lines known as circular functions, always 
bearing a fixed relationship to the radius of the circle 
and to each other, which gives rise to a number of 
terms, some of which, at least, it is desirable for the 
pattern cutter to understand. 

69. The Complement of an arc or of an angle is 
the difference between that arc or angle and a quad- 
rant. In Fig. 41, A D B is the complement of B D C, 
and vice vei'sa. 

70. The Supplement of an arc or of an angle is the 
difference between that arc or angle and a semicircle. 




Fig. 43. — Diagram Showing the Circular Functions 
of the Arc A H or Angle A C H. 

In Fig. 42, B D C is the supplement of A D B, and 
vice versa. 

71. The Sine of an arc is a straight line drawn 
from one extremity perpendicular to a radius drawn to 
the other extremity of the arc. (H B, Fig. 43.) 

72. The Co-Sine of an arc is the sine of the com- 
plement of that arc. H K, Fig. 43, is the sine of the 
arc A H. 

73. The Tangent of an Arc is a line which touches 
the arc at one extremity, and is terminated by a line 
passing from the center of the circle through the other 
extremity of the arc. In Fig. 43, A E is the tangent 
of A H or of the angle A C H. 

74. The Co-Tangent of an arc is the tangent of 
the complement. Thus F G, Fig. 43, is the co-tan- 
gent of the arc A H. 



Terms and Definitions. 



75. The Secant of an arc is a straight line drawn 
from the center of a circle through one extremity of 
that arc and prolonged to meet a tangent to the other 
extremity of the arc. (E C, Fig 43.) 

76. The Co-Secant of an arc or angle is the secant 
of the complement of that arc or angle, as F C, Fig. 

43. 

77. The Versed Sine of an arc is that part of the 
radius intercepted between the sine and the circumfer- 
ence. (A B, Fig. 43.) 

78. An Ellipse is an oval-shaped curve (Fig. 44), 



focus, and the straight line (C D) the directrix. In this 
fio-ure any point, as N or M, is equally distant from 
F and the nearest point in C D, as H or K. (See defi- 
nition 113.) 

80. A Hyperbola (A B, Fig. 46) is a curve from any 
point in which, if two straight lines be drawn to two 
fixed points, their difference shall always be the same. 
Thus, the difference between E G and G L is H L. 
and the difference between E F and F L is B L. II L 
and B L are equal. The two fixed points, E and L, 
are called foci. (See definition 113.) 





Fig. 44. — -4k Ellipse. 



Fig. 46.— A Hyperbola. 






Fig. 47. — Evolute and Involute. 



Fig. 45, — A Parabola. 



Fig. 4S.—A Trian- 
gular Prism. 



from any point in which, if straight lines be drawn to 
two fixed points within the curve, their sum will be 
always the same. These two points are called foci (F 
and H). The line A B, passing through the foci, is 
called the major or transverse axis. The line E Gr, per- 
pendicular to the middle of the major axis, and extend- 
ing from one side of the figure to the other, is called 
the minor or conjugate axis. There are various other 
definitions of the ellipse besides the one given here, 
dependent upon the means employed for drawing it, 
which will be fully explained at the proper place 
among the problems. (See definition 113.) 

79. A Parabola (A B, Fig. 45) is a curve in which 
any point is equally distant from a certain fixed point 
and a straight line. The fixed poinr (F) is called the 



81. An Evolute is a circle or other curve from 
which another curve, called the involute or evolutent, is 
described by the aid of a thread gradually unwound 
from it. (Fig. 47.) 

82. An Involute is a curve traced by the end of a. 
string wound upon another curve or unwound from it, 
(Fig. 471) (See also Prob. 84, Chapter IV.) 

SOLIDS. 

83. A Solid has length, breadth and thickness. 

84. A Prism is a solid of which the ends are equal, 
similar and parallel straight-sided figures, and of which 
the other faces are parallelograms. 

85. A Triangular Prism is one whose bases or ends 
are triangles. (Fig. 48.) 



s 



Tfie New Metal Worker Pattern Booh. 



86. A Quadrangular Prism is one whose bases or 
ends are quadrilaterals. (Fig. 49.) 

87. A Pentagonal Prism is one whose bases or ends 
are pentagons. (Fig. 50.) 

88. A Hexagonal Prism is one whose bases or ends 
are hexagons. (Fig. 51.) 

89. A Cube is a prism of which all the faces are 
squares. (Fig. 52.) 

90. A Cylinder, or properly speaking a Circular Cylin- 
der, is a round solid of uniform diameter, of which the 
ends or bases are equal and parallel circles. (Fig. 53.) 






Fig. p.— A Quad- 
rangular Prism. 



Fig. 50.— A Pent- 
agonal Prism. 



Fig. 51.— A Hex- 
agonal Prism. 






Fig. 52. — A Cube. Fig. 53.— A Cylinder. Fig 54 —A Cone. 
Apex 




Base 

Fig 55.— A Right 
Cone. 





Fig. 56 —An Oblique Fig. 57.— A Trun- 
or Scalene Cone. cated Cone. 



91. An Elliptical Cylinder is one whose bases are 
■ellipses. 

92. A Right Cylinder is one whose curved surface 
is perpendicular to its bases. 

93. An Oblique Cylinder is one whose curved sur- 
face is inclined to its base. 

94. A Cone is a round solid with a circle for its 
base, and tapering uniformly to a point at the top 
called the apex. (Fig 54.) 

95. A Right Cone is one in which the perpendicular 
let fall from the vertex upon the base passes through 
the center of the base. This perpendicular is then 
called the axis of the cone. (Fig. 55.) 

96. An Obliqe Cone or Scalene Cone is one in which 
the axis is inclined to the plane of its base. (Fig. 56.) 



97. A Truncated Cone is one whose apex is cut of! 
by a plane parallel to its base. (Fig. 57.) This figure 
is also called a frustum of a cone. A pyramid may also 
be truncated. (See Figs. 69 and 70 and definition 
112.) 

98. A Pyramid is a solid having a straight-sided 
base and triangular sides terminating in one point or 
apex. Pyramids are distinguished as triangular, quad- 
rcmgular, pentagonal, hexagonal, etc., according as the 
base has three sides, four sides, five sides, six sides, 
etc. (Figs. 58, 59 and 60.) 






Fig 5S.—A Trian- Fig. 59. — .4 Quadran- 
gular Pyramid. gular Pyramid. 



Fig. eu.—Aii Octag- 
onal Pyramid. 






Fig. 61.— A Right Fig. 62.— Altitude Fig 63 —Altitude 
Pyramid. of a Cone. of a Pyramid. 



/ 

V-l 




Fig. 65 — Altitude 
of a Cylinder. 



Fig. 66.— A Sphere, 
or Globe. 



Fig. 64.— Altitude 
of a Prism. 

99. A Right Pyramid is one whose base is a regular 
polygon, and in which the perpendicular let fall from 
the apex upon the base passes through the center of 
the base. This perpendicular is then called the axis 
of the pyramid. (Fig. 61.) 

100. The Altitude of a pyramid or cone is the 
length of the perpendicular let fall from the apex to 
the plane of the base. The altitude of a prism or 
cylinder is the distance between its two bases or ends, 
and is measured by a line drawn from a point in one 
base perpendicular to the plane of the other. (Figs. 
56, 62, 63, 64 and 65.) 

101. The Slant hight of a pyramid is the distance 
from its apex to the middle of one of its sides at the 
base. The slant hight of a cone is the distance 



and Definitions. 



from its apex to any point in the circumference of its 
base. 

102. A Sphere or Globe is a solid bounded by a 
uniformly curved surface, any point of which, is equally 
distant from a point within the sphere called the center. 
(Fig. 66.) 

103. A Polyhedron is a solid bounded by plane 
figures. There are five regular polyhedrons, viz. : 

104. A Tetrahedron is a solid bounded by four 
equilateral triangles. It is one form of triangular 
pyramid. (Fig. 67.) 

105. A Hexahedron is a solid bounded by six 
squares. The common name for this solid is cube, 
which see. (Fig. 52.) 



to each other as to appear as though one passes through 
the other. The intersection of their surfaces forms 
the basis of the greater part of the problems of Chap. 
VI. 

112. The Frustum of a Cone or Frustum of a Pyra- 
mid is that portion of the original solid which remains 
after the apex has been cut away upon a plane parallel 
to the base. (Figs. 57, 69 and 70.) When the cut- 
ting plane is oblique to the base of the solid they are 
spoken of as oblique frustums. 

113. A Conic Section is a curved line formed by the 
intersection of a cone and a plane. The different 
conic sections are the triangle, the circle, the ellipse, 
the parabola and the hyperbola. When the cutting 






Fig. 67.— A Tetra- 
hedron. 



Fig. 68.— An Octa- 
hedron. 



Fig. 69. — Frustum 
of a Scalene Cone. 





A C 

Fig. 72. — A Cone Cut by a Plane Parallel 
to One of Its Sides. 





Fig. 73.— A Cone Cut by a Plane Which 
Makes an Angle with the Base Greater 
than the Angle Formed by the Side. 





Fig. 70. — Frustum, 
of a Pyramid. 



Fig. 71. — .4 Cone Cut by a Plane Obliquely 
through Its Opposite Sides. 



106. The Octahedron is a solid bounded by eight 
equilateral triangles. (Fig. 68.) 

107. The Dodecahedron is a solid bounded by 
twelve pentagons. 

108. The Icosahedron is a solid bounded by twenty 
equilateral triangles. 

109. An Axis is a straight line, passing through a 
body on which it revolves, or may be supposed to 
revolve. (Figs. 55 and 61.) 

110. By the Envelope of a solid is meant the sur- 
face which encases or surrounds it, as the envelope of 
a cone. 

111. Intersection of Solids is a term used to describe 
the condition of solids which are so joined and fitted 



plane passes obliquely through its opposite sides the 
resulting figure is called an ellipse. (Fig. 71.) (An 
ellipse is also an oblique section through a cylinder.) 
When a cone is cut by a plane parallel to one of its' 
sides, the resulting figure is a parabola. Thus in Fig. 
72 the cutting plane A B is parallel to the side of 
the cone C D. See definition 79. When the cuttino- 
plane makes a greater angle with the base than the side 
of the cone makes, or when it passes vertically through 
the cone to one side of the axis, the resulting figure is 
a hyperbola. Thus in Fig. 73 the angle A B C is greater 
than the angle ADE. See definition SO. The para- 
bola and hyperbola resemble each other, both beino- 
incomplete figures, with arms extending indefinitely. 



10 



Tlie Neio Metal Worker Pattern Book. 



The ellipse is a complete figure, but of varying pro- 
portions, as the cutting plane is inclined more or less. 
114. Concave means hollowed or curved inward, 




Fig. 74.— Sections of Curved Surfaces. 

said of the interior of an arched surface or curved line 
in opposition to convex. (Fig. 74.) 

115. A Convex surface is one that is curved out- 
ward, that is regularly protuberant or bulging, when 
viewed from without. The opposite of convex is con- 
cave. (Fig. 74.) 

ARCHITECTURAL TERMS. 

116. The term Cornice is ordinarily used to desig- 
nate any molded projection or collection of moldings 
which finishes or crowns the part to which it is affixed. 
The term in this sense is applicable in all styles of 
architecture. In classical architecture, however, it is con- 
fined to the upper division of the entablature, the whole 



has been omitted. The names of parts given in the 
illustration are such as are generally understood by' 
architects and cornice makers. The cornice of clas- 
sical architecture may contain simply a bed mold, 
planceer and crown mold, or it may contain, in addi- 
tion, a dentil course or a modillion course, or both. 

117. The Entablature was used by the ancients to 
finish a wall or colonnade (more especially the latter), 
and consisted of three parts, the cornice, the frieze and 
the architrave. (Fig. 75.) 

IIS. The Architrave, the lower division of the 
entablature, was in reality a lintel used to span the 
space between the columns, but its form was main- 
tained when used above a wall. In modern imitations 
of the antique styles the molded portion is frequently 
used without the fascias, in which case it is commonly 
known as the foot mold. (Fig. 75.) The term arch- 
itrave is also used to designate the molding and fascias 
running around an arch or a window opening. 

119. The Frieze, the middle division of the entab- 
lature, is really a continuation of the wall surface to 
add hight and effect to the building, and was originally 
intended for the display of symbols, inscriptions, orna- 
ments, &c, appropriate to the use of the building of 
which it was a part. It is sometimes treated very 
plainly and sometimes receives considerable ornamen- 
tation, being subdivided into panels or enriched by 



ENTABLATURE < 



Q-lBED COURSE 



MODILLION 
COURSE 




Fig. 75.— The Entablature and Its Parts. 



of which, according to modern ideas, might be considered 
as a cornice. The distinction is shown in Fig. 75, 
which shows an entablature of a design adaptable to 
sheet metal construction, and in which all enrichment 



scrolls, etc. The terms plain frieze, designating a 
frieze devoid of ornamentation, and frieze-piece or frieze- 
panel, are used to designate one of the parts of which a 
frieze is constructed. (Fig. 75.) 



Terms and Definitions. 



11 



120. Arcb. The curved top of an opening in a wall. 
The arch of masonry is constructed of separate blocks 
and is supported only at the extremities. The joint 
lines between the blocks are disposed in the direc- 



plan, designed as a support for an entablature. It con- 
sists of three parts : a base, a shaft and a capital. 

122. An Engaged Column is a column placed 
against the face of a wall or other surface, from which 





; isrr— 1 




r 


i 




i ' 






i 






M 1 








1 




i 


1 




| 1 


i 




1 


1 






i / — «« 



Fig. 76. — A Semicircular Arch. 



Fig. 77.— A Pointed Arch. 



Fig. SO.— A Pilaster. 



tion of radii of the curve, thus enabling the arch to 
support the weight of the wall above the opening. 
"When in classical designs its face is finished with 
moldings their proper profile is that of an architrave. 
(Fig. 75.) The level lines at which the curve of the 
arch begins are called the springing lines. Sometimes 
the lower stones of the arch rise vertically a short 
distance from the supports before the springing lines 
are reached, in which case the arch is said to be stilted. 



it projects one-half or more than one-half its diam- 
eter. 

123. A Pilaster differs from a column in that it is 
square in plan instead of round and is usually engaged 
within a wall. (Fig. 80.) 

124. Pedastal. A structure designed to support 
a column, statue, vase or other object. It is by some 
described as the foot of a column, but is, properly 
speaking, not a part of it. It consists of three 





Fig. 81. — An Ayigular Pediment. 



A 




Fig. 78. — A Moresque Arch. 



Fig. 79.— A Flat Arch. 



Fig. 82. — A Segmental Pediment. 



The stones composing the arch are called the voussoirs, 
and the middle or top stone is called the keystone. The 
supports below the ends of the arch are called imposts. 

Arches are usually semicircular (Fig. 76), semi- 
elliptical, segmental, pointed (Fig. 77) or Moresque (horse- 
shoe) (Fig. 78) in shape, according to the style of archi- 
tecture with which they are used. 

The top of an opening may be perfectly level and 
yet composed of wedge-shaped blocks so combined as 
to be self-supporting, in which case it is called a. fiat 
arch. (Fig. 79.) 

121. A Column is a vertical shaft or pillar round in 



parts, a base, a middle portion cubical in shape called 
a die and a cap or cornice. It is also used as a finish 
at the ends of a balustrade course. 




Fig. 8$. — A Broken Pediment. 



125. A Pediment is a triangular or segmental orna- 
mental facing over a portico, door, window, etc. (Figs. 
81, 82 and 83. 



12 



The New Metal Worker Pattern Book. 



126. A Broken Pediment is one, either in the form 
of a gable or a segment, which is cut away in its 
central portion for the purpose of ornamentation. 
(Fig. 83.) 

127. A Gable is the vertical triangular end of a 
house or other building, from the cornice or eaves to 
the top. 

128. A Lintel Cornice is a cornice above or some- 
times including a lintel. This term is very generally 
used to designate the cornice used above the first story 
of stores. (Fig. 81.) 

129. A Deck Cornice or Deck Molding is the cornice 




Fig. 84. — A Lintel Cornice. 




Fig. 85. — A Bracket. 

or molding used to finish the edge of a flat roof where 
it joins a steeper portion. 

130. A Bracket, as used in sheet metal work, is 
simply an ornament of the cornice. Brackets in stone 
architecture were originally used as supports of the 
parts coming above them. Hence modern architecture 
has kept up that idea in their designs. (Fig. 85.) 

131. Modillions are also cornice ornaments, and 
differ from brackets only in general shape. (Fig. 86.) 
While a bracket has more depth than projection, modil- 
lions have more projection than depth. 

132. A Dentil is a cornice ornament smaller than 
a modillion, which in shape usually represents a solid 
with plain, rectangular face and sides. Dentils are 
never used singly, but in courses, the spaces between 
them being less than their face width. (Fig. 76.) 



133. A Corbel is a modified form of bracket. It is 
used to terminate the lower parts of window caps, and 
also forms the support for arches, etc. , in gothic forms. 





Fig. 86. — -4 Modillion. 



Fig. 87.— A Head Block. 



131. A Head Block or Truss is a large terminal 
bracket in a cornice, projecting sufficiently to receive 
all the moldings against its side, thus forming a finish 
to the end of the cornice. (Fig. 87.) 

135. A Stop Block is a block-shaped structure, vari- 
ously ornamented, which is placed above the end 





Fig. 88.— A Stop Block. 



Fig. 89.— A Pinnacle. 



bracket in a cornice, and which projects far enough to 
receive against its side the various moldings occurrina: 
above the brackets, forming an end finish. (Fig. 88.) 
136. A Pinnacle is a slender turret or part of a 
building elevated above the main building. A small 
spire. (Fig. 89.) 



Ten 



id Definitions. 



13 



137. A Finial is an ornament variously designed, 
placed at the apex of a pediment, gable, spire or 
roof. 

138. Capital. — The upper member or head of a col- 
umn or pilaster. It may vary in character according to 
the style of architecture with which it is employed, from 
a few simple projecting moldings around the top of the 
column to an elaborately foliated ornament. The lower- 

^ABACUS 
|VOLUT£ 




NEXrt 
MOLD 



Fig, 90.— Capitals, 



most mold is called the neck mold and the uppermost 
member sustaining the weight of the lintel or arch 
above is called the abacus. (Fig. 90.) 

139. Panel. — Asunken compartment having molded 
edges used to ornament a plane surface, as a frieze ceil- 
ing, planceer or tympanum. A panel may, however, 
be raised instead of sunken. 

The margin or space between the sides of the panel 
and the edges of the surface in which it is placed is 
usually made equal all around and is called the stile. 

140. A Volute is a spiral scroll used as the principal 
ornament of a capital and is placed under the corners 
of the abacus. For method of drawing the volute see 
Probs. 81 and 82, Chap. IV. 

141. A Molding is an assemblage of forms project- 
ing beyond the wall, column or surface to which it is 
affixed. (See first part of Chap. V.) 

142. Crown Molding is the term applied to the upper 
or projecting member of a cornice. (Fig. 75.) 

143. Planceer or Plancher is the ceiling or under 
side of the projecting part of a cornice. (Fig. 75.) 

141. The Bed Moldings of a cornice are those mold- 
ings forming the lower division of the cornice proper, 
and which are made up of the bed course, modillion 
course and dentil course. (Fig. 75.) 

145. The Bed Course is the upper division of the 
bed moldings, the part with which the bracket heads 
and modillion heads ordinarily correspond, and against 
which they miter. (Fig. 75.) 

146. The Modillion Course of a cornice embraces 
the modillions and all the moldings which are imme- 
diately back of and below them. The plain surface 
lying back of or between the modillions is called in 



sheet metal work the modillion band, and the molding 
immediately below them the modillion molding. (Fig. 
75.) 

147. The Dentil Course of a cornice embraces the 
dentils and all the moldings to which the dentils are 
attached as ornaments, comprising the dentil band and 
dentil molding. (Fig. 75.) 

148. Foot Molding is the common term used to 
designate the lower molding in a cornice. It is fre- 
quently in this connection used in the sense of archi- 
trave. (Fig. 75.) 

149. A Bracket Molding, also called bracket head, is 
the molding around the upper part of a bracket, and 
which generally members with the bed molding, against 
which it finishes. (Fig. 75.) 

150. A Gable Molding is an inclined molding which 
is used in the finish of a gable. 

151. A Ridge Molding is a molding used to cap or 
finish a ridge. It is also called a ridge capping or sim- 
ply ridging. 

152. A Hip Molding is a molding used to protect 
and finish the hips or angles of a roof. It is very fre- 
quently included in the more general term ridging. 

153. A Fascia is a plain band or surface below a 
molding, or, in other words, the unornamented face of a 
portion of a cornice or architrave. (Fig. 75.) 

154. A Fillet is a narrow plain member of a mold- 
ing used to finish or separate the different forms 
(a a a a Fig. 75 are fillets.) 

155. A Drip is a downward projecting member 
in a cornice or in a molding, used to throw the water 
off from the other parts. (Fig. 75.) 

156. Soffit is the term applied to the under side 
of a projecting molding, cornice or arch. 

157. A Sink is a depression in the face of a piece 




I 

-s— 

Fig. 91. — Stays. 

of work or in a plain surface. (See face of bracket, 
Fig. 85, side of modillion, Fig. 86.) 

158. Incised Work is a style of ornamentation 
consisting of fine members and irregular lines, sunken 
or cut into a plain surface. (See side of bracket Fig. 
85.) 

159. The Stay of a molding is its shape or profile 
cut in sheet metal. (Fig. 91.) 



14 



TJie New Metal Worker Pattern Book. 



160. Rake Moldings are those which are inclined, 
as in a gable or pediment ; since to miter a rake mold- 
ing with a level return under certain conditions neces- 
sitates a change or modification of profile in one or 
the other of the moldings to rake means to make such 
change of profile. 

161. A Raked Molding, therefore, is a term describ- 




Fig. 92. — Elevation of a House. 

ing a molding of which the profile is a modification of 
some other profile. 

1G2. A Raked Profile or Raked Stay describes the 
profile or stay which has been derived from another 
profile or stay, by certain established rules, in a process 
like that of mitering a horizontal and inclined molding 
together. 

163. The Normal Profile or Normal Stay is the 
original profile or stay from which the raked profile or 
stay has been derived. 




Fig. 93.— Plan of a House. 

164. A Flange is a projecting edge by which a 
piece is strengthened or fastened to anything. 

165. A Hip is the external angle formed by the 
meeting of two sloping sides or skirts of a roof which 
have their wall plates running in different directions. 

DRAFTING TERMS. 

166. Projection is that department of geometrical 
drawing which treats of the drawing of elevations, 
plans, sections and perspective views. There are three 



kinds of projections in general use, viz. : Orthographic, 
Isometric and Perspective. Chapter III is devoted to 
an explanation of the principles of orthographic pro- 
jection. 




H'ig. 9f. — Section of House on Line 
A B of Plan and Elevation, 

167. An Elevation is a geometrical projection of a 
building or other object on a plane perpendicular to 
the horizon. (Fig. 92.) 

168. A Plan is the representation of the parts as 
they would appear if cut by a horizontal plane. (Fig. 
93.) 

169. A Section is a view of the object as it would 
appear if cut in two by a given vertical or horizontal 
plane. (Fig. 94.) In the one case the resulting view 
is called a vertical section, and in the other a horizontal 




Fig. 95. — Perspective View of House. 

section. Oblique sections are representations of objects 
cut at various angles. 

170. A Perspective is a representation of a build- 
ing or other object upon a plane surface as it would 
appear if viewed from a particular point. (Fig. 95.) 

171. A Detail Drawing or Working Drawing is a 
drawing, commonly full size, for the use of mechanics 
in constructing work. 

172. A Scale Drawing is one made of some scale 
less than full size. 



Terms and Definitions. 



15 



173. A Miter is a joint in a molding, or between 
two pieces not moldings, at any angle. 

174. A Butt Miter is the term applied to the cut 
made upon the end of a molding to fit it against 
another molding or against a surface. 

175. A Gable Miter is the name applied to the 
miter either at the peak or at the foot of the moldings 
of a gable or pediment. 

176. A Rake Miter is a miter between two mold- 
ings, one of which has undergone a modification of 
profile to admit of the joint being made. 

177. Square Miter is the common term for a 
joint at right angles, or at 90°. 



178. An Octagon Miter is a miter joint between 
two sides of a regular octagon, or between any two 
pieces at an angle of 135°. 

179. An Inside Miter indicates a joint at an inte- 
rior or re-entrant angle. 

180. An Outside Miter is a joint at an exterior 
angle. 

181. The Development of a surface is the process 
of finding, from a drawing of a rounded form, a shape 
or pattern upon a flat surface which, when cut out and 
bent or formed as indicated by that drawing, will con- 
stitute its envelope ; or, in other words, the stretching 
out flat of a surface shown by a drawing to be curved. 



Alphabetical List. 



In the following list all words are arranged in alphabetical order, the figure following each referring 
to the number of the definition in the list preceding : 



Abacus 138 

Acute Angle 14 

Altitude 35. I0 ° 

Angle 12 

Angle, Right 13 

Angle, Acute 14 

Angle, Obtuse 15 

Apex 31 

Arc 59 

Arch 120 

Architrave 118 

Axis 109 

Base 32, 121, 124 

Bed Course 145 

Bed Moldings 144 

Bracket 130 

Bracket Molding 149 

Broken Pediment 126 

Butt Miter 174 

Capital 138 

Center 54 

Chord 60 

Circle 52 

Circumference 53 

Circumscribed 66 

Column 121 

Complement 69 

Concave 114 

Concentric 64 

Cone 94 

Cone, Oblique or Scalene.. 96 



Cone, Truncated 97 

Conic Section 113 

Convex 115 

Corbel 133 

Cornice 116 

Co-Sine 72 

Co-Secant 76 

Co-Tangent 74 

Crown Molding 142 

Cube 89 

Cylinder 90 

Cylinder, Elliptical 91 

Cylinder, Oblique 93 

Cylinder, Right 92 

Decagon 48 

Deck Cornice 129 

Degree 68 

Dentil 132 

Dentil Course 147 

Detail Drawing 171 

Development 181 

Diagonal 51 

Diameter 56 

Die 124 

Dodecagon 49 

Dodecahedron 107 

Drip 155 

Elevation 167 

Ellipse 78 

Elliptical Cylinder 91 

Engaged Column 121 



Entablature 117 

Envelope no 

Evolute 81 

Eccentric 65 

Fascia 153 

Fillet 154 

Finial 137 

Flange 164 

Foot Molding 148, 118 

Frieze 119 

Frustum 112 

Gable 127 

Gable Miter 175 

Gable Molding 150 

Geometry 1 

Globe 102 

Head Block 134 

Heptagon 46 

Hexagon 45 

Hexahedron 105 

Hip 165 

Hip Molding 152 

Hyperbola 80 

Hypothenuse 30 

Icosahedron 10S 

Impost 120 

Incised Work 158 

Inscribed 66, 67 

Inside Miter 179 

Intersection of Solids in 

Involute 82 



Keystone 120 

Line 4 

Line, Curved 6 

Line, Horizontal 8 

Line, Inclined or Oblique.. 10 

Lines, Parallel 7 

Line, Perpendicular n 

Line, Straight 5 

Line, Vertical 9 

Lintel Cornice 128 

Lozenge 41 

Miter ■ 173 

Modillion 131 

Modillion Course 146 

Molding 141 

Neckmold 138 

Normal Profile 163 

Oblique Cone 96 

Oblique Cylinder 93 

Obtuse Angle 15 

Octagon 47 

Octagon Miter 178 

Octahedron 106 

Outside Miter 180 

Panel 139 

Parabola 79 

Parallelogram 39 

Pedestal 124 

Pediment 125 

Pediment, Broken 126 

Pentagon 44 



16 



Vie New Metal Worker Pattern Book. 



Perimeter... 50 

Perspective 170 

Pilaster 123 

Pinnacle 136 

Plan 168 

Planceer 143 

Plane 17 

Plane Figure 20 

Point 3 

Polygon 22 

Polyhedron .... 103 

Prism 84 

Prism, Hexagonal 88 

Prism, Pentagonal 87 

Prism, Quadrangular 86 

Prism, Triangular 85 

Projection 166 

Pyramid 98 

Pyramid, Right 99 

Quadrant 62 

Quadrilateral ............ 36 



Radius 55 

Rake 160 

Rake Miter 176 

Rake Moldings 160 

Raked Molding 161 

Raked Profile 162 

Rectangle 42 

Rectilinear Figure 21 

Rhomboid 40 

Rhombus 41 

Ridge Molding 151 

Right Angle 13 

Right Cone 95 

Right Cylinder 92 

Right Line 5 

Right Pyramid 99 

Scale Drawing 172 

Scalene Cone 96 

Scalene Triangle 26 

Secant 75 

Section 169 



Sector 61 

Segment 58 

Semicircle 57 

Shaft 121 

Sides of a Triangle 22 

Sink 157 

Sine 71 

Slant Hight 101 

Soffit 156 

Solid 83 

Sphere 102 

Springing Line 120 

Square 43 

Square Miter 177 

Stay 159 

Stile 139 

Stilted 120 

Stop Block 135 

Supplement 70 

Surface 16 

Surface, Double Curved.. , 19 



Surface, Single Curved.. .. 18 

Tangent 63 

Tangent of an Arc 73 

Tetrahedron 104 

Trapezium 37 

Trapezoid 38 

Triangle 23 

Triangle, Acute-Angled. . . 28 

'triangle, Equilateral 24 

Triangle, Isosceles 25 

Triangle, Obtuse- Angled. . 29 

Triangle, Right- Angled. . . 27 

Triangle, Scalene 26 

Truncated Cone 97 

Truncated Pyramid 97 

Truss 134 

Versed Sine 77 

Vertex 34 

Volute 140 

Voussoir 120 

Working Drawing 171 



CHAPTER II. 



Brawling Tools amid Materials. 



To the person about to begin a new occupation 
the first consideration is, what tools and materials does 
he need ? In the following description of the appli- 
ances, tools and materials likely to be of service to the 
pattern cutter in the class of work in which he is sup- 
posed to be the most interested, the description is limited 
to articles of general use. Those who are interested 
in drawing tools and materials upon a broader basis 
than here presented are referred to special treatises on 
drawing and to the catalogues of manufacturers and 
dealers in drawing materials and drawing instruments. 

Drafting: Tables.— A drafting table suitable for a 
jobbing shop should be about five feet in length and 
three to four feet in width. It is better to have a 
table somewhat too large, than to have one so small 
that it is frequently inadequate for work that comes 
in. In flight the table should be such that the drafts- 
man, as he stands up, may not be compelled to stoop 
to his work. While for some reasons it is desirable 
that the table should be fixed upon a strong frame and 
legs, for convenience such tables are generally made 
portable. Two horses are used for supports and a 
movable drawing board for the top. A shallow drawer 
is hung by cleats fastened to the under side, and is ar- 
ranged for pulling either way. Sometimes horizontal 
jfieces are fastened to the legs of the horses, and a 
shelf or shelves are formed by laying boards upon 
them. Fig. 96 shows such a table as is here described. 
When properly made, using heavy rather than light 
material, such a table is quite solid and substantial, 
and when not in use can be packed away into a very 
small space. 

For cornice makers' use, a table similar in con- 
struction to the one described and illustrated (Fig. 96) 
is well adapted. Its dimensions, however, consider- 
ing the extremes of work that are likely to arise, 
should be twelve to fourteen feet in length by about 
five feet in breadth. Three horses are necessary, and 
two drawers may be suspended. For very large work, 



one draftsman or pattern cutter will require the whole 
table, but for ordinary work, such as window caps, 
cornices, etc., two men can work at it without interfer- 
ing with or incommoding each other. 

Various woods may be used for drawing tables, 
but white pine is the cheapest and best for the pur- 
pose. Inch and one-half to two-inch stuff will be 
found economical, as it allows for frequent redressing 
— made necessary by pricking in the process of pattern 
cutting. Narrow stuff, tongued and grooved together 
or joined by glue, is preferable to wide plank, as it is 




Fig. 96.— Drafting Table. 

less liable to warp. Rods run through the table edge- 
ways, as shown in Fig. 96, are desirable for drawing 
the parts together and holding them in one compact 
piece. The nut and washer are sunk into the edge of 
the table, a socket wrench being used to operate them. 
A drafting table should be an accurate rectangle 
— that is, every corner should be a right angle, and 
the opposite sides should be parallel. The edges 
should be exactly straight throughout their length. 
Methods of testing drafting tables and drawing boards, 
with reference to these points, are given below. The 
usual way of adjusting a table or board to make it ac- 
curate is to plane off its edges as required. But this 
is a task less simple than it appears. It requires the 
nicest skill and accuracy to render it at all satisfactory. 



IS 



The New Metal Worker Pattern Book. 



When it is remembered that no matter how well sea- 
soned the lumber employed may be the table will be 
affected by even slight changes in the atmosphere, it 
is apparent that dressing off the edges with a plane, 
under certain circumstances, might be constantly re- 
quired. For great accuracy, adjustable metal strips 
may be fastened to the edges of the table in such a 
manner that, by simply turning a few screws, any 
variation in the table may be compensated. This ar- 
rangement may be accomplished in the following man- 
ner : The edge of the table on all sides is cut away so 
as to allow a bar of steel, say one-eighth or one-six- 
teenth of an inch thick and abdut an inch wide, to lie 
in the cutting, so that its surface is even with the face 
of the table, with its outer edge projecting somewhat 
beyond the edge of the table. Slotted holes are made 
in the table through which bolts with heads counter- 
sunk into the metal are passed for holding the steel 
strips. A washer and nut are used on the under side 
of the table. The adjustment required is, of course, 
very slight. The edge of the metal projecting slightly, 
as described, is well adapted for receiving the head of 
the T-square, rendering the use of that instrument 
more satisfactory than when it is used against the plane 
edge of the table, even if equally accurate. 

Drawing Boards. — The principal difference between 
a drafting table and a drawing board is in the size. The 
same general requirements in point of accuracy, etc., 
are necessary in each. Convenient sizes of tables for 
various uses have been mentioned, but to point out 
sizes of boards for different purposes is not so easy a 
matter, their application being far more extended and 
their use more general. A drawing board may be made 
of any required size, from the smallest for which such 
an article is adapted up to the extreme limit con- 
sistent with convenience in handling. In the larger 
sizes the general features of construction noted under 
drafting tallies are entirely applicable, save that thinner 
material should be used in order to reduce the weight. 
In small sizes there is a choice between several different 
modes of construction, two or three of which will be 
described, although boards of almost any required con- 
struction can be purchased ordinarily of dealers in 
drawing tools and materials at lower prices than they 
can lie made. However, it is very convenient, in many 
cases, to have boards made to order, and therefore de- 
tailed descriptions of good constructions are desirable. 
Any carpenter or cabinet maker should be able to do 
the work. 

In Fig, 97 is shown a very common form of draw- 



ing board, consisting of a pine wood top with hard- 
wood ledges. The ledges are put on by means of a 
dovetail, tapering probably one-half inch in the width 
of the board, so that while allowing entire freedom for 




Fig. 97. — Drawing Board, with Ledges. 

seasoning there is no danger of cracking the board, 
and they may be driven tight as required. Where it 
is desirable to use screws in the ledges they are passed 
through slotted holes furnished with a metallic bush- 
ing. 

In Fig. 98 is shown a still simpler form of board, 
which is adapted only for the smallest sizes. Hard- 




Fig. 98. — Drawing Board, with Tongued and Grooved Cleats. 

wood strips are tongued and grooved onto the ends to 
prevent warping, as shown in the engraving. By using 
strips of wood thicker than the board, keeping their 
upper surfaces flush with the surface of the board, it 
may be constructed so as to have the advantage of 
ledges on the under side equivalent to those shown in 
Fig. 97. 

Fig. 99 shows a construction which, while being 




Fig. 99. — Bottom View of Drawing Board, with Grooved Back and 
Cleats Attached with Slotted Holes. 



somewhat more expensive than the others, is un- 
doubtedly much better. It is made of strips of pine 
wood, glued together to make the required width. A 



Drawing Tools and Materials. 



19 



pair of hard-wood cleats is screwed to the back, the 
screws passing through the cleats in oblong slots with 
brass bushings, which fit closely under the heads and 
yet allow the screws to move freely when drawn by the 
shrinkage of the board. To overcome the tendency of 
the surface to warp, a series of grooves are sunk in half 
the thickness of the board over the entire back. To 
make the working edges perfectly smooth, allowing an 
easy movement with the "T-square, a strip of hard-wood 
is let into the end of the board. The strip is afterward 
sawn apart at about every inch, to admit of contraction. 
In the construction of such boards additional advantage 
is obtained by putting the heart side of each piece of 
wood to the surface. 

As pattern cutting is nothing if not accurate, it is 
a matter of the utmost importance that the drawing 
board or table should be perfectly rectangular. If each 
angle is a right angle — if its opposite sides are exactly 
parallel — the T- s q uar e: may be used at will from any 
portion of it with satisfactory results. If the board is 
accurate the drawing will be accurate If the board is 
not accurate the drawing can only be made accurate at 




Fig. 100 —Testing the Sides of a Drawing Board. 

the cost of extra trouble and care. While it is easy to 
get a board approximately correct by ordinary means, 
one or two simple tests will serve to point out inac- 
curacies for correction which by ordinary means would 
pass unnoticed. For such tests a T-square and an or- 
dinary two-foot steel square that are exactly correct 
will be required. 

Having made the opposite sides and ends of the 
board as nearly accurate as possible, place the head of 
the T-square against one side, as shown in Fig. 100, 
and with a hard pencil sharpened to a chisel edge, or 
with the blade of a knife, scribe a fine line across the 
board. Then carrying the T-square to the opposite 
side of the board, as shown by the dotted lines, bring 
the edge of the blade to the line just scribed and see 
that it exactly coincides throughout its length with the 
line. Repeat this operation at frequent intervals along 
the edges of the board, both at the sides and ends. 
Remove any small inaccuracies on the edges by means 



of a file or fine sand paper folded over a block of 
wood. Careful work in this manner will produce very 
satisfactory results. 

A means of testing a board with reference to the 
accuracy of the corners is shown in Fig. 101. A car- 
penter's try-square or an ordinary steel square used 




Fig. 101. — Testing the Corner of a Drawing Board. 

upon the corners does not ordinarily reach far enough 
in either direction to satisfactorily determine that the 
adjacent end and side are perpendicular to each other: 
hence it is desirable to obtain some kind of a test with 
reference to this point from the middle portions of the 
edges. With the head of the T-square placed against 
one side of the board draw a fine line, as indicated by 
the dotted line in the engraving, and from one end draw 
a second line in the same manner. If the side and end 
are at right angles the two lines will coincide with the 
arms of a square when placed as shown in the engrav- 
ing. Repeat this operation for each of the corners. 
The two methods above described for testing drawing 
boards, especially when used together, cannot fail to 
enable any one to obtain a board as nearly accurate as 
it is possible to make it. Modifications of the methods 
here given, and based upon the same principles, will 
suggest themselves to any one who will give the mat- 
ter careful thought. 

Straight-Edges. — In connection with every set of 
drawing instruments there should be one or more 



Fig. 102.— Straight-Edge. 

straight-edges. If nothing but pencil or pen lines are 
to be made upon paper, those of hard-wood or hard 
rubber will answer very well; but if lines are to be 
drawn upon metal, steel is the only satisfactory ma- 
terial. The length of the straight-edge must be de- 
termined by the work to be done, but a safe rule is to 
have it somewhere near the length of the table or 
board. Of course this is out of the question in cor- 
nice work, where tables are frequently upward of 



20 



TJie New Metal Worker Pattern Book. 



twelve feet in length. In such cases the size of the 
material to be cut determines this matter. If iron 96 
inches long is used, the straight-edge, for convenience, 
should not be less than Si feet. If shorter iron is 
regularly used, a shorter straight-edge will answer. In 
cornice work, two and even three different lengths are 
found advantageous. The longest might be as just 
described; a second might be about four feet in 
length and made proportionately lighter, while the 
smallest might be two feet and still lighter than the 
four-foot size. Instead of the latter, however, the 
long arm of the common steel square serves a good 
purpose. 

For tinners' use in general jobbing shops, a three- 
foot straight-edge in many cases, and a four-foot one 
in a few instances, will be found very convenient. 
Some mechanics desire their straight-edges graduated, 
the same as a steel square, into inches and fractions. 
There is, however, no special advantage in this ; it 
adds considerably to the cost, without rendering the 
tool more useful. 

A hole should be provided in one end of the 
straight-edge for hanging up. It should always be 
suspended when not in use, as in that position it is not 
liable to receive injury. 

It is almost superfluous to add that straight-edges 
must be absolutely accurate, for if inaccurate they 
would belie their name. A simple and convenient 
method of testing straight-edges is to place two of 
them together by their edges, or a single one against 
the edge of a square, and see if light passes between 
them. If no space is to be observed between the 
edges it is satisfactory evidence that they are as nearly 
straight as they can be made by ordinary appliances. 
In addition to having the edges straight it is also 
necessary to have the two sides parallel. 

T-Squares. — With this instrument, as with almost 
all drawing instruments, there is the ehoice of various 
qualities, sizes and kinds, and selection must be made 
with reference to the kind of work that is to be per- 
formed. Whatever quality may be chosen, the de- 
sirable features of a T-square are strict accuracy in ail 
respects, and a thin, flat blade that will lie close to the 
paper. For most purposes a fixed head, as shown in 
Fig. 103, is preferable. For drawings in which a 
great number of parallel oblique lines are required, and 
particularly where a small size T-square can be used, a 
swivel head, as shown in Fig. 104, is sometimes de- 
sirable. The objectionable featm-e about a swivel head 
is the difficulty of obtaining positive adjustment. 



When made in the ordinary manner, and depending 
upon the friction of the nut of a small bolt for holding 
the head in place, it is almost impossible to obtain a 
bearing that can be depended upon during even a 
simple operation. In practice it is found to be far less 
trouble to work from a straight-edge — properly placed 
across the board and weighted down or otherwise held 



n 



Fig ]03.—l-Square with Fixed Head. 

in place — by means of a triangle or set-square, as 
greater accuracy is thus assured. 

In point of materials, probably a "[-square having a 
walnut head and maple blade is as satisfactory as any. 
This kind is the cheapest and is generally considered 
the best for practical purposes. A good article, but 
of higher price, consists of a walnut head with a hard- 
wood blade, edged with some other kind of wood. 
Still another variety has a mahogany blade edged with 
ebony. T-squares constructed with cast-iron head — 
open work finished by japanning — with nickel-plated 
steel blade, are also to be had from dealers. They are 
also made with a hard rubber blade, of which Fig. 104 
is an illustration. The liability to fracture, however, 
by dropping necessitates the greatest care in use; 
otherwise hard rubber makes a very desirable article 
and is the favorite material with many draftsmen. 

As to size, T-squares should be selected with 
reference to the use to be made of them. Generally, 
the blade should be a very little less in length than the 
width of the table or board upon which it is to be 




Fig. 104.—1-Square with Fixed and Surtvel Head. 

used. Where a large board or a table is used it will 
be found economical to have two instruments of dif- 
ferent sizes. 

The Steel Square. — One of the most useful tools in 
connection with the pattern cutter's outfit is an ordinary 
steel square. The divisions upon it concern him much 
less than its accuracy. He seldom requires other 
divisions than inches and eighths of an inch ; therefore 
in selection the principal point to be considered is that 



Drawing Tools and Materials. 



21 



of accuracy. The finish, however, is a matter not to 
be overlooked. Since a nickel-plated square costs but 
a trifling advance upon the plain article, it is cheaper 
in the long run to have the plated tool. 

A convenient method of testing the correctness of 
the outside of a square, and one which can be used 
at the time and place of purchase, is illustrated in 



Fig. 105. — Testing the Exterior Angle of a Steel Square. 

Fig. 105. Two squares are placed against each other 
and against a straight-edge, or against the arm of a 
third square. If the edges touch throughout, the 
squares may be considered correct. 

Having procured a square which is accurate upon 
the outside, the correctness of the inside of another 
square may be proven, as shown in Fig. 106. Place 
one square within the other, as shown. If the edges 



i ■ i ■ i ■ i ■ i ■ i ' i ■ i ' i ■ i • i ■ 



T* 



1 I ■ I ' I ■ J_ - 



■'.■!■ I I.I. I i-l.l.l.i.l.i.l. 1,1 I 1,1,1,1,1, l,l,l,lil, l,l,l,l,l,lil,U,n 



i ■ i • i ' i ■ i ' i • i ■ i ' -i ■ t ■ i 4 

I.I.I.1.I.I.I.I.I , 1,1,1,1.1,1,1,1, i.i.i.i.u.i.i.ij.iri 

Fig. 106.— Testing the Interior Angle of a Steel Square. 

fit together tightly and uniformly throughout, the 
square may be considered entirely satisfactory. 

An accurate square is especially desirable, as it 
affords the readiest means of testing the T-square and 
the drawing table and beard, as elsewhere described. 
The greatest care should be given, therefore, to the 
selection of a square. For all ordinary purposes the 



two-foot size is most desirable. In some cases the 
one-foot size is better suited. Many pattern cutters 
on cornice work like to have both sizes at their com- 
mand, making use of them interchangeably, according 
to the nature of the work to be done. 

Triangles, or Set Squares.— In the selection of tri- 
angles, the draftsman has the choice in material be- 




Fig. 107.— Open Hard Rubber Triangle or Set Square, 45 x 45 x 90 

Degrees. 



tween pear wood ; mahogany, ebony lined ; hard rub- 
ber; German silver, and steel, silver or nickel plated. 
In style he has the choice between open work, of the 
form shown in Fig. 107, and the solid, as in Fig. 
108. In shape, the two kinds which are adapted to 
the pattern cutter's use are shown in Figs. 107 and 
108, the latter being described as 30, 60 and 90 de- 
grees, or 30 by 60 degrees, and the former as 45, 45 
and 90 degrees, or simply 45 degrees. The special 
uses of each of these two tools are shown in the 
chapter on Geometrical Problems (Chap. IV). In size, 
the pattern cutter requires large rather than small 




Fig. lOS.—Hard Wood Triangle or Set Square, SO x GO x 90 Degrees. 

ones. If he can have two sizes of each, the smaller 
should measure from 4 to 6 inches on the side, and 
the larger from 10 to 12 inches; but if only a single 
size is to be had, one having dimensions intermediate 
to those named will be found the most serviceable. 

The value of a triangle, for whatever purpose used, 
depends on its accuracy. Particularly is this to be said 
of the right angle, which is used more than either of 
the others. A method of testing the accuracy of the 
right angle is shown in Fig. 109. Draw the line A B 



22 



The New Metal Worker Pattern Book. 



with an accurate ruler or straight-edge. Place the 
right angle of a triangle near the center of this line, as 
shown bj D C B, and make one of the edges coincide 
with the line, and then draw the line D C against the 




■B— 



Fig. 109.— Testing the Right Angle of a Triangle. 

other edge. Turn the triangle into the position indi- 
cated by D C A. If it is found that the sides agree 
with A C and C D, it is proof that the angle is a right 
angle and that the sides are straight. 

Besides the kinds of triangles described above, a 
fair article can be made by the mechanic from sheet 
zinc or of heavy tin. Care must, however, be taken 
in cutting to obtain the greatest possible accuracy. For 
many of the purposes for which a large size 45 degree 



a profile line — is called spacers, as illustrated and de- 
scribed below. 

A pair of compasses consists of the parts shown 
in Fig. 110, being the instrument proper with detach- 
able points, and extras comprising a needle point, a 
pencil point, a pen and a lengthening bar, all as shown 
to the left. In selection, care should be given to the 
workmanship ; notice whether the parts fit together 
neatly and without lost motion, and whether the joint 
works tightly and yet without too great friction. A 
good German silver instrument, although quite ex- 
pensive at the outset, will be found the cheapest in 
the end. A pencil point of the kind shown in our 
engraving is to be preferred over the old style which 
clamps a common pencil to the leg. The latter is not 
nearly so convenient and is far less accurate. 

Of dividers there are two general kinds, the plain 
dividers, as shown in Fig. Ill, and the hair-spring 
dividers, as shown in Fig. 112. The latter differ from 
the former simply in the fact of having a fine spring 
and a joint in one leg, the movement being controlled 
by the screw shown at the right. In this way, after 




Fig. 110. — Compasses with 
Interchangeable Parts. 



Fig. 111.— Plain 
Dividers. 




Fig. 



112.— Hair-Spring 
Dividers. 



Fig. US. — Steel Spring 
Spacers. 



triangle would be used the steel square is available, 
but as the line of the hypothenuse is lacking, it can- 
not be considered a substitute. 

Compasses and Dividers. — The term compasses is 
applied to those tools, of various sizes and descriptions, 
which.hold a pencil and pen in one leg, and are used for 
drawing circles, while dividers are those tools which, 
while of the same general form as compasses, have 
both legs ending in fixed points, and are used for 
measuring spaces. A special form of dividers — used 
exclusively for setting off spaces, as in the divisions of 



the instrument has been set approximately to the dis- 
tance desired, the adjustable leg is moved, by means 
of the screw, either in or out, as may be required, 
thus making the greatest accuracy of spacing possible. 
Both instruments are found desirable in an ordinary 
set of tools. The plain dividers will naturally be used 
for larger and less particular work, while the hair- 
spring dividers will be used in the finer parts. It fre- 
quently happens that two pairs of dividers, set to dif- 
ferent spaces, are convenient to have at the same time. 
A pair of spacers, shown in Fig. 113, is almost 



Drawing Tools and Materials. 



23 



indispensable in a pattern cutter's outfit. He will find 
advantageous use for this tool, even though possessing 
both pairs of dividers described above. In size they 
are made less than that of the dividers. The points 
should be needle-like in their fineness, and should be 
capable of adjustment to within a very small distance 
of each other. It is sometimes desirable to divide a 
given profile into spaces of an eighth of an inch. The 
spacers should be capable of this, as well as adapted 
to spaces of three-quarters of an inch, without being 
too loose. As will be seen from the engraving, this 
instrument is arranged for minute variations in ad- 
justment. 

Beam Compasses and Trammels — In Fig. 114 is 
shown a set of beam compasses, together with a portion 
of the wooden rod or beam on which they are used. 
The latter, as will be seen by the section drawn to one 
side (A), is in the shape of a T. This form has con- 
siderable strength and rigidity, while at the same time 
it is not clumsy or heavy. Beam compasses are pro- 
vided with extra points for pencil and ink work, as 
shown. While the general adjustment is effected by 
means of the clamp against the wood, minute varia- 
tions are made by the screw shifting one of the points, 
as shown. This instrument is quite delicate and when 
in good order is very accurate. It should be used only 
for fine work on paper and never for scribing on metal. 

A coarser instrument, and one especially designed 
for use upon metal, is shown in Fig. 115 and is called 




Fig. 114. — Beam Compasses. 

a trammel. It is to be remarked in this connection 
that the name trammel, by common usage, is applied 
to this instrument and also to a device for drawing 
ellipses, which will be found described at another place. 
There are various forms of this instrument, all being 
the same in principle. The engraving shows a form 



in common use. A heavier stick is used with it than 
with the beam compasses, and no other adjustment is 
provided than that which is afforded by clamping 
against the stick. In the illustration a carrier at thr 




Fig 115. — Trammel. 

side is shown in which a pencil may be placed. Some 
trammels are arranged in such a manner that either of 
the points may be detached and a pencil substituted. 

A trammel, by careful management, can be made 
to describe very accurate curves, and hence can be used 
in place of the beam compasses in many instances. 
For all coarse work it is to be preferred to the beam 
compasses. It is useful for all short sweeps upon 
sheets of metal, but for curves of a very long radius a 
strip of sheet iron or a piece of wire will be found of 
more practical service than even this tool. 

The length of rods for both beam compasses and 
trammels, up to certain limits, is determined by the 
nature of the work to be done. The extreme length 
is determined by the strength and rigidity of the rod 
itself. It is usually convenient to have two rods for 
each instrument, one about 3|- or i feet in length and 
the other considerably longer — as long as the strength 
of material will admit. In the case of the trammel, 
by means of a simple clamping device, or, in lieu ci 
better, by use of common wrapping twine, the rods 
may be spliced when unusual length is required ; but 
a strip of sheet iron or a piece of fine wire forms 
a better radius, under such circumstances, than the 
rod. 



The New Metal Worker Pattern Book. 



The Protractor is an instrument for laying down 
and measuring angles upon paper. The instrument 
consists of a semicircle of thin metal or horn, as rep- 
resented in Fig. 116, the circumference of which is 
divided into ISO equal parts or degrees. The princi- 
ples upon which the protractor is constructed and used 
are clearly explained in the chapter on Terms and Defi- 
nitions (Def. 68 "Degree "). The methods of employing 
it in the construction of geometrical figures are shown 




Fig. 116. — Semicircii'ar Protractor. 

in Chapter IV among the problems. For purposes of 
accuracy, a large protractor is to be preferred to a 
small size, because in the former fractions of a degree 
are indicated. 

While a number of geometrical problems are con- 
veniently solved by the use of this instrument, it is 
not one that is specially adapted to the pattern cutter's 
use. All the problems which are solved by it can be 
worked out by other accurate and expeditious methods, 
which, in most cases, are preferable. It is one of the 
instruments, however, included in almost every 
case of instruments sold, and the student will find 
it advantageous to become thoroughly familiar with 
it, whether in practice he employs it 
or not. 

Besides the semicircular form of the ,. ..,,<,. 

protractor shown, corresponding lines and E 

divisions to those upon it are sometimes 
put upon some of the varieties of scales 
in use, as shown in Fig. 120. 

Scales. — Many of the drawings from which tne 
pattern cutter works — that is, from which he gets di- 
mensions, etc., — are what are called scale drawings, 
being some specified fraction of the full size of the 
object represented. Architects' elevations and floor 
plans are very generally made either ■§■ or \ inch 
to the foot, or, in other words, -^ or -^ full size. 
Scale details are also employed quite extensively by 
architects, scales in very common use for the purpose 



being 1-J- inches to the foot and 3 inches to the foot, 
or, in other words, -§- and J full size respectively. It 
is essential that the pattern cutter should be familiar 
with the various scales in common use, that he may 
be able to work from any of them on demand. Sev 
eral of the scales are easily read, by means of the com- 
mon rule, as, for example, 3 inches to the foot, in 
which each quarter inch on the rule becomes one inch 
of the scale ; also, 1-| inches to the foot, in which 
each eighth of an inch on the rule becomes an inch of 
the scale ; and, likewise, f inch to the foot, in which 
each sixteenth of an inch on the rule becomes an inch 
of the scale. However, other scales besides these are 
occasionally required, which are not easily read from 
the common rule, and sometimes special scales are 
used, which are not shown on the instruments, espe- 
cially calculated for the purpose. Accordingly, it is 
sometimes necessary for the pattern cutter to construct 
his own scale. 

The method of constructing a scale of 1 inch to 
the foot is illustrated in Fig. 117, in which the di- 
visions are made by feet, inches and half inches. In 
constructing such scales, it is usual to set off the di- 
visions representing feet in one direction (say to the 
right) from a point marked 0, while the divisions for 
inches and fractions thereof are set off the opposite- 
way (or to the left from 0) as shown in the illustra- 
tion. In using the scale, measurements are made by 
placing one point of the dividers at the number of 
feet required ; the other point can then be moved to 
the other side of the to the required number of 
inches, thus embracing the entire number of feet and 
inches between the points of the dividers. 

Besides scales of the kind just described, which 



ches 

S 5 4 3 t J 



Feet 



ffi 



Fig. 117. — Plain Scale (1 inch to the Foot ) 



are termed plain divided scales, there are in common 
use what aye known as diagonal scales, an illustration 
of one of which is shown in Fig. 118. The scale rep- 
resented is that of 1|- inches to the foot. The left- 
hand unit of division has been divided Idj means of 
the vertical lines into 12 equal parts, representing 
inches. In width the scale is divided into 8 equal 
parts by means of the parallel lines running its entire 
length. Next the diagonal lines are drawn, as shown. 



Drawing Tools and Materials. 



25 



By a moment's inspection it will be seen that, by 
means of these diagonal lines, one-eighth of an inch 
and multiples thereof are shown on the several hori- 
zontal lines. A distance equal to the space from A to 
B, as marked on the scale, is read (first at the right for 
feet) 2 feet (then to the left for inches by means of 



A flat scale is also manufactured in both boxwood 
and ivory. Fewer scales or divisions can be put upon 
it than upon the triangular scale, yet for certain pur- 
poses it is to be preferred to the latter. There are 
less divisions to perplex the eye in hunting out just 
what is required, and accordingly, there is less lia- 



1 


! 1 1 ID 






Inches 
s 












Feet 2 


V 

8 












I 








| 










/ 












I 






I 




1 








| 
















! 


I 








8 
















' 






; 








\ 




















.' 










% 


















i 


' 
























I 








I 








3 


' 


- 








i 



















IB II 10 9 & 7 6 5 * 3 2 I 



Fig. 118. — Diagonal Scale (!■>£ inches to the foot). 



the vertical lines figured both at top and bottom) 6 
inches (and last by means of the diagonal line, figured 
at the end of the scale, for fractions) and three-eighths. 
The top and bottom lines of the scale measure feet and 
inches only. The other horizontal lines measure feet, 
inches and fractions of an inch, each horizontal line 
having its own particular fraction, as shown. Such 
scales aPe frequently quite useful, as greater accuracy 
is obtained and, as the reader will see, may be con- 
structed by any one to any unit of measurement, and 
divided by the number of horizontal lines into any de- 
sired fractions. 

A scale in common use, and known as the tri- 
angular scale, is shown in Fig. 119. The shape of 
this scale, which is indicated by the name, and which 




Fig 119 — Triangular Boxwood Scale. 

is also shown in the cut, presents three sides for divi- 
sion. By dividing each of these through the center 



lengthways by a groove, as shown, six spaces for 
divisions are obtained, and by running the scales in 
pairs — that is, taking two scales, one of which is twice 
the size of the other, and commencing with the unit 
at opposite ends — the number of scales which may be 
put upon one of these instruments is increased to 
twelve. This article, which may be had in either 
boxwood, ivory or plated metal, and of 6, 12, 18 or 
24 inches in length, is probably the most desirable for 
general use of any sold. 



bility to error in its use. However, the limited num- 
ber of scales which it contains greatly restricts its 
usefulness. 

Fig. 120 shows another form of the flat scale, in 
quite common use in the past, but now virtually dis- 
carded in favor of more convenient dimensions and 
shapes. This scale combines with the various divi- 
sions of an inch the divisions of the protractor, as 
shown around the margin. The fact that the divisions 
of an inch for purposes of a scale are located in the 
middle of the instrument, away from the edge, which 
makes it necessary to take off all measurement with 
the dividers, renders the article awkward for use, and 



\ v. v \ 



X Vo \ ^TT 



1-^ I \w .W.-^ ft/o 7/0 »/0 



/ / / / / ,/ y 



•■ '.--■' ' -"> ■ ■/'" s 



4 ■> 



12 13 14 15 16 



Fig. ISO. — Flat Scale with Divisions of the Protractor on the Margins. 

the arrangement of the divisions of the circle, on the 
margins, is less satisfactory for use than the circular 
protractor. 

Lead f encils. — Various qualities of pencils are sold, 
some at much lower prices than others, but, all things 
considered, in this as in other cases, the best are the 
cheapest. The leading brands are made in two grades 
or qualities. The ordinary grades employ numbers, 
1, 2, 3, etc., to indicate hardness of lead, No. 1 being 
the softest, and No. 5 being the hardest in common 
use. A finer grade of pencils, known as poligrades, is 
marked by letters, commencing at the softest with B B, 



26 



The New Metal Worker Pattern Book. 



and ending at the hardest with H H H H H H, while 
other makes of pencils are marked by systems peculiar 
to their manufacturer. The draftsman has the choice 
of round or hexagon shape in all except the finest 
grades, the latter being made exclusively hexagon. 
Whatever kind of pencil the draftsman or mechanic 
uses, he will require different numbers for different 
purposes. For working drawings, full-sized details, 
etc., on manila paper, a No. 3 (or F) is quite satis- 
factory. Some like a little harder lead, and therefore 
prefer a No. 4 (or H). For lettering and writing in 
connection with drawings upon manila or ordinary 
detail paper, a No. 2 (H B) is usually chosen. For 
fine lines, as in developing a. miter, in which the great- 
est possible accuracy is required, a No. 5 is very gen- 
erally used, although many pattern cutters prefer the 
finer grade for this purpose and use a H I H H H. 

The quality and accuracy of drawings depend; in 
a considerable measure, upon the manner in which 
pencils are sharpened. A pencil used for making fine 




Fig. 121.— Two Views of Pencil Sharpened to a Chisel Point, 

straight lines, as, for instance, in the various opera- 
tions of pattern cutting, should be sharpened to a chisel 
point, as illustrated in Fig. 121, Pencils for general 
work away from the edges of the J-square, triangle, 
etc., should be sharpened to a round point, as shown 
in Fig. 122. It facilitates work and it is quite eco- 
nomical to have several pencils at command, sharpened 
in different ways for different purposes. Where for 
any reason only one pencil of a kind can be had, both 
ends may be sharpened, one to a chisel point and the 
other to a round point. 

For keeping a good point upon a pencil, a piece 
of fine sand paper or emery paper, glued upon a piece 
of wood, will be found very serviceable. A flat file, 
mill-saw cut, is also useful for the same purpose. 
Sharpen the pencil with a knife, so far as the wood 
part is concerned, and then shape the lead as required 
upon the file or sand paper. 

Drawing Pens. — Although most of the pattern 
cutter's work is done with the pencil, there occasion- 
ally arise circumstances under which the use of ink is 
desirable. Tracings of parts of drawings are frequently 



required which can be better made with ink than with 
pencil. 

The drawing pen or ruling pen, as illustrated in 
Fig. 123, is used for drawing straight lines. The 
drawing pen, whether as a separate instrument or as 
an attachment to compasses or beam compasses for 
drawing curved lines, consists of two blades with steel 
points, fixed to a handle. The blades are so curved 
that a sufficient cavity is left between them for ink 
when the points meet close together or nearly so. The 
space between the points is regulated by means of the 
screw shown in the engraving, so as to draw lines of 
any required thickness. One of the blades is provided 
with a joint, so that, by taking out the screw, the. 
blades may be completely opened and the points readily 
cleaned after use. The ink is put between the blades 
with a common pen, or sometimes by a small hair 
brush. Tn using the drawing pen it should be slightly 
inclined in the direction of the line to be drawn, and 
should be kept uniformly close to the ruler or straight- 



Fig. 122. — Pencil Sliarpened to a Round Point, 



Fig. 123. — Ruling Pen. 

edge during the whole operation of drawing a line, 
but not so close as to prevent both points from touch- 
ing the paper equally. 

Keeping the blades of the pen clean is essential 
to good work. If the draftsman is careless in this par- 
ticular, the ink will soon corrode the points to such an 
extent that it will be impossible to draw fine lines. 

Pens will gradually wear away, and in course of 
time they require dressing. To dress up the tips of 
the blades of a pen, since they are generally worn 
unequally by customary usage, is a matter of some 
nicety. A small oil stone is most convenient for use 
in the operation. The points should be screwed into 
contact in the first place, and passed along the stone, 
turning upon the point in a directly perpendicular plane 
until they acquire an identical profile. Next they are 
to be unscrewed and examined to ascertain the parts 
of unequal thickness around the nib. The blades are 
then to be laid separately upon their backs upon the 
stone, and rubbed down at the points until they are 
brought up to an edge of uniform fineness. It is well 
to screw them together again and pa.ss them over the 



Drawing Tools and Materials. 



27 



stone once or twice more to bring up any fault and to 
retouch them also at the outer and inner side of each 
blade to remove barbs or frazing, and finally to draw 
them across the palm of the hand. 

India Ink. — For tracings, and for some kinds of 
drawings, which the pattern cutter is obliged to make 
occasionally, India ink is much better than the pencil, 
which is used for the greater part of his work. Care 
is to be exercised in the selection of ink, as poor grades 
are sold as well as good ones. Some little skill is 
required in dissolving or mixing it for use. 

India ink is sold in cakes or sticks, of a variety 
of shapes. It is prepared for use by rubbing the end 
of the stick upon the surface of a ground glass, or of a 
porcelain slab or dish, in a very small quantity of 
water, until the mixture is sufficiently thick to produce 
a black line as it flows from the point of the ruling 
pen. The quality of ink may generally be determined 
by the price. The common size sticks are about 3 
inches long. Inferior grades can be bought as low as 
40 cents per stick, while a good quality is worth $1.50 
to $2 per stick, and the very best is still higher. How- 
ever, except in the hands of a responsible and expe- 
rienced dealer, this method of judging is hardly 
satisfactory. To a certain extent ink may be judged 
by the brands upon it, although in the case of the 
higher qualities the brands frequently change, so that 
this test may not be infallible. The quality of India 
ink is quite apparent the moment it is used. The best 
is entirely free from grit and sediment, is not musky, 
and has a soft feel when wetted and smoothed. The 
color of the lines may also be used as a test of quality. 
With a 230or ink it is impossible to make a black line. 
It will be brown or irregular in color and will present 
an irregular edge, as though broken or ragged, while 
an ink of satisfactory quality will produce a clean line, 
whether drawn very fine or quite coarse. 

Various shaped cups, slabs and dishes are in use 
for mixing and containing India ink. In many re- 
spects they are like those used for mixing and holding- 
water colors.' Indeed, in many cases the same articles 
are employed. The engraving (Fig. 124) shows what 
is termed an India ink slab, with three holes and one 
slant. This article is in common use among draftsmen 
and serves a satisfactory purpose. In order to retard 
evaporation, a kind of saucers, in sets, is frequently 
used, so constructed that one piece will form a cover to 
the other, and which are known in the trade as cabinet 
sets or cabinet saucers. They are from 2-J- to 3£ 
inches in diameter and come in sets of six. In the 



absence of ware especially designed for the purpose, 
India ink can be satisfactorily mixed in and used from 
an ordinary saucer or plate of small size. The articles 
made especially for it, however, are convenient, and 




Front with Cover On. 




Top with Cover Off. 
Fig. 124 —India Ink Slab. 

in facilitating the care and economical use of the ink 
are well worth the small price they cost. 

Several makes of liquid drawing ink are also to 
be had, which possess the advantage of being always 
ready for use, thus doing away with the rubbing 
process. The ink costs about 25 cents a bottle, 
keeps well, and will answer almost every purpose 
quite as well as the stick ink. 

Thumb Ta.ks or Drawing Pins, both names being 
in common use, are made of a variety of sizes, ranging 
from those with heads one-quarter of an inch in diam- 



Fig. 125. — Thumb Tacks, or Drawing Pins, 

eter up to eleven-sixteenths of an inch in diameter. 
They are likewise to be had of various grades and 
qualities. The best for general use are those of Ger- 
man silver, about three-eighths to five-eighths of an 
inch in diameter, and with steel points screwed in and 
riveted Those which have the points riveted only 
are of the second quality. The heads should be flat> to 
allow the T-square to pass over them readily. In the 



28 



The New Metal Worker Pattern Book. 



annexed cut, Fig. 125, are shown an assortment of 
kinds and sizes. Those which are beveled upon their 
upper edges are preferable to those which are beveled 
underneath. 

A Box Of Instruments.— Fig. 126 shows a box of in- 
struments of medium grade, as made up and sold by 
the trade generally. While it contains some pieces 
that the pattern cutter has no use for, it also contains 
the principal tools he requires, all put together in com- 
pact shape, and in a convenient manner for keeping 
the instruments clean and in good order. The tray of 
the box lifts out, there being a space underneath it in 
which may be placed odd tools, pencils, etc. Tools 
luay be selected, as recpiired, of most of the large 
dealers in drawing instruments. It will be found ad- 
vantageous to the pattern cutter to buy his instruments 
singly as he requires them, as by so doing he will get 
only what he recpiires for use, and will probably secure 




»:- Si 



A 



Fig. 126. — .4 Box of Instruments. 



a better quality in the tools. After he has made his 
selection, a box properly fitted and lined should be 
provided for them and can be obtained at a small cost, 
or made if desirable. 

India Rubber. — A good rubber with which to erase 
erroneous lines is indispensable in the pattern cutter's 
outfit. The several pencil manufacturers have put 
their brands upon rubber as well as upon pencils, and 
satisfactory quality can be had from any of them. The 
shape is somewhat a matter of choice, flat cakes being 
the most used. A very soft rubber is not so well 
adapted to erasing on detail paper as the harder varie- 
ties, but is to be preferred for use in fine drawings on 
good quality paper. 

Paper. — The principal paper that the pattern cutter 
has anything to do with is known as brown detail paper, 
or manila detail paper. It can be bought of almost 



any width, from 30 inches tip to 54 inches, in rolls of 
50 to 100 pounds each. It is ordinarily sold in the 
roll by the pound, but can be bought at retail by the 
yard, although at a higher figure. There are different 
thicknesses of the same quality. Some dealers indi- 
cate them by arbitrary marks, as XX, XXX, XXXX; 
others by numbers 1, 2, 3; and still others as thin, 
medium and thick. The most desirable paper for the 
pattern cutter's use is one which combines several good 
qualities. It should be just as thin as is consistent 
with strength. A thick paper, like a stiff card, breaks 
when folded or bent short, and is, therefore, objection- 
able. The paper should be very strong and tough, as 
the requirements in use are quite severe. The surface 
should be very even and smooth, yet not so glossy as 
to be unsuited to the use of hard pencils. It should 
be hard rather than soft and should be of such a 
texture as to withstand repeated erasures in the same 
spot without damage to the surface. 

"White drawing paper, which the pattern cutter 
has occasionally to use in connection with his work, 
can be had of almost every conceivable grade and in a 
variety of sizes. The very best quality, and the kinds 
suited for the finest drawings, come in sheets exclu- 
sivety, although the cheaper kinds are also made in 
the shape of sheets as well as in rolls. White drawing 
paper in rolls can be bought of different widths, rang- 
ing from 36 to 54 inches, and from a very thin grade 
up to a very heavy article, and of various surfaces. It 
is sold by the pound, in rolls ranging from 30 to 40 
pounds each, and also at retail by the yard. A kind 
known as eggshell is generally preferred by architec- 
tural draftsmen. 

Drawing paper in sheets is sold by the quire, and 
at retail by the single sheet. The sizes are generally 
indicated by names which have been applied to them. 
The following are some of the terms in common use, 
with the dimensions which they represent placed op- 
posite : 

17 



Cap 13 

Demy 15 

Medium 17 

Royal • 19 

Super Royal..... 19 x 27 

Imperial 22 x 30 



x 
x 
x 

X 



20 
22 
24 



Elephant 23 x 28 

Atlas...... 26 x 34 

Columbier... 23 x 35 

Double Elephant. 27 x 40 

Antiquarian .. 31 x 53 

Emperor 48 x 68 



Still another set of terms is used in designating French 
drawing papers. Different qualities of paper, both as 
regards thickness, texture and surface, can be had of 
any of the sizes above named. 



Drawing Tools and Materials 



29 



Tracing: Paper and Tracing Cloth.— The pattern 
cutter has frequent use for tracing paper, and a good 
article, which combines strength, transparency and 
suitable surface, is very desirable. Tracing paper is 
sold both in sheets, in size to correspond to the draw- 
ing papers above described, and in rolls, to correspond 
in width to the roll drawing paper. It is usually 
priced by the quire and by the roll, although single 
sheets or single yards are to be obtained at retail. 
The rolls, according to the kinds, contain from 20 to 
30 yards. There are various manufacturers of this 
article, but it is usually sold upon its merits, rather 
than by any brand or trade-mark. Tracing cloth, or 
tracing linen, is used in place of tracing paper where 
great strength, and durability are required. This 



article comes exclusively in rolls, ranging in width 
from 18 to 42 inches. There are generally 24 yards 
to the roll, and prices are made according to the 
width, or, in other words, according to the superficial 
contents of the roll. Two grades are usually sold, the 
first being glazed on both sides and suitable only for 
ink work, and the second on but one side, the other 
being left dull, rendering it suitable for pencil marks. 
Upon general principles, pencil marks are not satis- 
factory upon cloth, even upon the quality specially 
prepared with reference to them. It is but a very 
little more labor or expense to use ink, and a much 
more presentable and usable drawing is made. Tracing 
paper may be used satisfactorily with either pencil or 
pen. 



¥ 



-^e> 



CHAPTER III, 



Linear Brawl egc 



In the production of all great constructive works 
the drawing plays a most important part. If a piece 
of machinery, a ship, an aqueduct or a temple is to be 
built, verbal descriptions would be insufficient direc- 
tions to the workmen who are to perform the actual 
labor; drawings become a necessity, because a draw- 
ing tells exactly what is meant, where words would 
utterly fail. Therefore, to everybody connected with 
the constructive trades, to artisans in whatever field, 
the ability to read, if not to make, a drawing becomes 
a necessity ; and to those in positions ot authority the 
ability to make a drawing is the power to convey their 
ideas to others. That branch of drawing with which 
the pattern cutter has to deal is of a purely geomet- 
rical nature and is properly termed orthographic projec- 
tion. The term orthographic (signifying right line) is 
well applied because it exactly describes the nature of 
the work, as will be seen further on. 

The geometrical drawings made use of in repre- 
senting any constructive work, whether to a large or a 
small scale, are of three kinds — viz. : Elevations, 
sections and plans. The term diagram is sometimes 
used in connection with this class of drawing, but is 
not of a specific nature. It means a drawing of the 
simplest possible character, usually made to demon- 
strate a principle, and may partake of the properties of 
either of the above named drawings. 

An elevation, if the word were judged by its com- 
mon meaning, would be understood to show the bight 
of anything. It does this and more. It gives all the 
vertical and horizontal measurements which appear in 
the front, side or end which it represents. An elevation 
supposes the observer to be opposite to and on a level 
with all points at the same time, and is therefore an 
impossible view, according to the rules of pictorial art. 
Being always drawn to scale (including full size), it 
gives exact dimensions of hight and breadth at any 
part of the view, but furnishes no view of horizontal 
surfaces and no means of measuring distances to and 



from the observer, or in any oblique horizontal direc- 
tion. An elevation may be called front, side, end or 
rear, according to the relative dimensions of the object, 
one of whose faces it represents. Any elevation or 
vertical section gives two sets of dimensions — i. e., 
hight and horizontal distance, which lie parallel to the 
face which it shows. 

A section, as the word indicates, is a view of a 
cut or a view of what remains after certain portions 
have been cut away for the purpose of showing more 
clearly the interior construction. The idea of a verti- 
cal section can best be described by supposing that a 
wire stretched taut, or any perfectly straight blade, 
was passed vertically down through an object at a given 
distance from one of its ends or sides, indicated by a 
line in some other view or views, and the portion not 
wanted w r as removed. The view made of the section 
may properly include only the parts cut, or if made to 
include or show portions that would naturally appear 
by the removal of the parts, it would properly be called 
a sectional elevation. Sections may also be taken hori- 
zontally at any hight above the base or ground line, 
indicated by a line for that purpose upon one or more 
of the elevations. 

Horizontal sections are properly classed with plans. 
Vertical sections are known as longitudinal or transverse, 
according as they are taken through the long way of, 
or across, an object. Elevations or sections may also 
be constructed upon oblique planes when necessary to 
more fully show construction. 

Sections of small portions or members drawn to a 
large scale or full size are called profiles. They are ap- 
plied to continuous forms, as moldings, jambs, etc., 
and are drawn for the purpose of showing the peculiari- 
ties in form of the parts which they represent. 

The view which gives all the horizontal distances 
in whatever direction is called the plan. The name 
plan applies equally well to a horizontal section or to 
a top view. In the plan, as in sections and elevations, 



Linear Drawing. 



31 



the observer is supposed to be opposite to (i.e. , di- 
rectly above) all points at the same time. In idea it 
is the same as a map, the difference between the two 
terms being in the amount included in the view. 

In Fig. 128 is given an illustration of the various 
geometrical views of an object, placed in their proper 
relation one to another, showing the lines of projec- 
tion and the lines upon which the different sections 
are taken. A house placed upon a base has been se- 
lected as the most suitable object for purposes of 
explanation in the present case. It has been shown 
in diagrammatic form — that is, denuded of all cornices, 
trimmings or projecting parts — so as to demonstrate 
the principles of projection in the clearest manner 
possible. 

It rests with the designer to determine which of 
the views shall be drawn first, all depending upon the 
given facts or specifications in his possession. If a 
house is to be designed, it is most likely that the plan 
would be drawn first, as arrangement of rooms and 
amount of ground to be covered would be of the first 
importance. If a molding be the subject of the de- 
sign, the profile would be the view in which to first 
adjust the proportion of its parts. The method of 
deriving the elevation from the section or obtaining 
any one view from one or more other views is termed 
orthographic projection, because by it a system of 
parallel lines is made use of for the purpose of obtain- 
ing the same hight (or width, as the case may be) in 
corresponding parts in the different views. 

In this connection it is to be understood that each 
angle or limit of outline in a sectional view is the 
source of a right line in the elevation. In Fig. 127 is 
shown, at X, a sectional view or profile of a molding, 
which should be so drawn that all the faces or sur- 
faces supposed to be vertical shall lie vertically on the 
paper; that is, parallel to the sides of the drawing 
board. To project an elevation, Y, from this section, 
place the T-square so that the blade lies horizontal — 
that is, crossing the board from side to side — and bring 
it to the various angles A, B, C, etc., of the profile, 
drawing a line from each. The point E, though not 
an angle, is the lowest visible point or limit of that 
member of the mold when seen from the front, and is, 
therefore, entitled to representation in the elevation 
by a line. In like manner the point D, being the 
upper limit of a curve, is entitled to representation, 
but being so situated as to be invisible when viewed 
from a point in front of the mold, the line is properly 
made dotted. The lines of projection from the section 



to the elevation are also shown dotted in the engrav- 
ing. A vertical line terminates the elevation of the 
mold at the right or end nearest the section, while the 
absence of such a line at its left end indicates that it 
extends indefinitely in that direction. It would also 
be proper, upon that supposition, to finish the eleva- 
tion at the left with a broken line. 

Referring now to Fig. 128, it is most likely that 
the front elevation would be next drawn after the plan. 
For this purpose the plan should be so placed upon 
the board that the part representing the front should 
be turned toward the bottom of the board, in which 
position it appears to be turned toward the observer. 
Place the T-square so that the blade lies vertically 
upon the board — that is, crossing it from front to 
back — and bringing it to the different angles or 
points of the front side of the plan, draw a line ver- 
tically from each, through that portion of space upon 



At 




Fig. 127. — Elevation Projected from Section. 

the paper allotted to the elevation, all as shown by the 
dotted lines. Thus each point of the elevation comes 
directly over the point which represents it in the plan, 
and the horizontal distance across any part of the new 
elevation thus becomes exactly the same as that of the 
plan. The question of bights is here a matter of de- 
sign and is governed by specifications supplemented 
by the designer's judgment. With the plan and the 
front elevation complete the drawing of any other ele- 
vations or sections is entirely a matter of projection, 
except as new features might occur in those vieWs 
which would not appear in either of the views already 
drawn. 

If an elevation of the right side is about to be con- 
structed, lines would be projected horizontally to the 
right from every point in the front elevation of the ob- 
ject which would be visible when seen from the right 
side, thus locating all the bights in the new view. As 
the horizontal distances in this view must agree with 
distances from front to back on the plan, they may best 
be obtained by turning the plan (or so much of it as nec- 
essary to this view) one-quarter around to the right, so 



32 



TJie New Metal Worker Pattern Book. 



.0. 







h. 

UJ 



t 
I 



z 

< to. 



Dr 



33 



that the side of which the new elevation is to be drawn 
will be toward the bottom or near side of the board, 
as shown at G ; after which lines may be projected with 
the T-square from the points of the plan into the eleva- 
tion, intersecting with corresponding lines, as shown. 
The same result may be accomplished by projecting 
the lines to the right from the side of the plan, as 
shown in the top view, until they reach any line paral- 
lel to the side, as H I. From this line they may be 
carried around a quarter circle from any convenient 
center, as N, arriving at a horizontal line, 1ST M, and 
thence dropped downward, intersecting as before. 

It will thus be seen that the elevation of the right 
hand side of any object comes naturally at the right of 
the front elevation, and the left side elevation, at its 
left. This idea is best illustrated by supposing that 
the object in question be placed in a glass box of the 
dimensions of the base H I J K of the top view, and 
that the elevation of each side of the object be pro- 




LONG1TUD1NAL SECTION ON CD TRANSVERSE SECTION ON A-B 

Fig. 129. — Vertical Sections Derived from Fig. 128. 



jected upon the adjacent parallel side of the box at 
right angles to the same, and that afterward all the 
sides (supposing them to be hinged at the corners) be 
opened out into one plane, as shown by K L, H and 
P (the top face of the box being opened upward), 
thus displaying all the views in one plane as repre- 
sented by Fig. 128. 

This idea should not be carried so far as to open 
the bottom face of the box downward, because this 
would produce a plan as seen from below, which is 
never done except in the case of a design of a ceiling 
or soffit, when it should be spoken of as an inverted 
plan. 

In Fig. 129 the transverse section is shown at the 
right of the longitudinal section, because the view in 
it is from the right, or in the direction of the arrow in 
the longitudinal section, showing what would be seen 
if the house were cut in two on the line A B of the 
plan and the right hand portion removed. The longi- 



tudinal section is for the same reason placed at the left 
of the transverse section — that is, it is a view from the 
left of the house when placed in the position shown by 
the transverse section. From the foregoing it is to be 
understood, therefore, that when a view appears to the 
right of another it is supposed to show what would be 
seen when the object is viewed from the right hand 
end or side of what is shown in the other, the other 
(or front) view being at the same time a view of the 
left side of what is shown by the right side elevation. 

In this class of drawings various kinds of lines 
are used, each of which possesses a certain significance. 
The general outlines of the different views should 
be firm and strong enough to be distinctly visible, with- 
out being so broad as to leave any doubt as to the ex- 
act dimensions of the part shown when the rule is ap- 
plied for purposes of measurement. 

It is not always necessary that all the lines of 
projection should be shown. When shown they may 
appear as the finest possible continuous lines, 
or as dotted lines such as are shown in Figs. 
127, 128 and 129. Lines used in carrying 
points from a profile to a miter line, or from 
one line to another for any purpose, are really 
lines of projection, and for the pattern drafts- 
man's purposes it may be said that the finer 
they are drawn the greater the accuracy ob- 
tained (see Chapter II under the head of Lead 
Pencils). 

Dotted lines are also used to represent 
portions which are out of sight — that is, back 
of or underneath the other parts which constitute the 
view under consideration, but which it is necessary to 
show, as, for instance, a portion of the chimney in 
longitudinal section Fig. 129 and points D and F in 
the profile of the mold in Fig. 127. 

Dotted lines are also used to show a change of 
position or an alternate position of some part, as, for 
example, the lines L K and J L show that the side J K 
of the top view has been swung around on the point K 
until it occupies the position shown by L K, its ex- 
tremity J traversing the line J L. When it is neces- 
sary to use two kinds of dotted lines, those used for 
one purpose may be made in fine or short dots, while 
the others may be made a series of short dashes. 

Lines showing the part of a view through which a 

section is taken are composed of a series of dots and 

dashes, as shown by A B, C D, etc., in Fig. 128, and 

when further distinction is required may be made by 

gi- I two dots alternating with a short or long dash. 



34 



The New Metal Worker Pattern Book. 



When it is desirable to omit the drawing of a con- 
siderable portion of any view it is customary to termi- 
nate the incomplete side of such view by an irregular 
line, as shown above the plan G in Fig. 128. 

It is customary in all sectional views for the parts 
which are represented as being cut to be ruled or lined 
with lines running in an oblique direction, as in Fig. 
129. When the section comprises several different 
pieces lying adjacent to one another, each different 
part should be lined in a different direction. This rul- 
ing is understood to mean solidity. In Fig. 129 the 
walls and base in the different sections are represented 
as though made of some solid material, as wood or 
stone, and ruled accordingly. Where it is necessary 
to represent different kinds of material in the same sec- 
tion, different systems or kinds of lines may be used 
for the purpose. Thus solid and dotted lines may be 
used alternately, as in the base. Coarse and fine rul- 
ing, or stippling, may also be employed, according to 
the size of the part, or very small parts may be shown 
solid black, as window weights, piping or hinges. A 
heavy line is the only way that a thickness of metal 
can properly be shown in a section. In the case of a 
sectional view of a cornice or molding where nothing 
but the sheet iron appears, it is customary to make use 
of section lines close to the metal surface, but not to 
extend them clear across the space which should be 
filled if the moldings were of stone or other solid mate- 
rial. By this means a section may be distinguished 
from what might otherwise be taken for an elevation 
of a return. 

In the case of elaborate drawings prepared by an 
architect color is frequently resorted to as a means of 
showing the different materials as they appear in the 
sectional view, yellow or differing shades of brown 
being used for various kinds of wood, while blue is 
generally used for iron, gray for stone, red for brick, 
etc. In the case of drawings showing many different 
materials it is usual to place a legend in one corner of 
the drawing showing what each color or style of ruling 
indicates. 

It is always advisable to keep the different views, 
which it is necessary to construct, separate and dis- 
tinct from one another, drawing them as near to- 
gether as circumstances will permit, but never allow- 
ing one view to cover any part of the space upon the 
paper occupied by the' other view if it can be avoided. 
One notable exception to this rule is to be observed. 
It frequently occurs in drawing an elevation of a large 
surface, as a pediment or side of a bracket, that it is 



necessary to indicate that some part of it is recessed or 
raised, or that a certain edge is molded or chamfered, 
when it would not be necessary to construct an entire 
sectional view for this purpose alone. To this end it 
is customary to draw through such mold, chamfer or 
recess a small section, in which case, if the depression 
or mold runs horizontally, the section is turned to the 
right or left, according to convenience, or if it runs 
obliquely, it is turned in the direction the mold runs. 
In such a section the line which represents the plane 
surface also shows the direction of the cut across the 
mold or line upon which the section is taken. In Fig. 
130 is shown an elevation of a portion of a pediment, 
in which a small section, A B C, is introduced to show 
the profiles of the moldings. The line B C, which 




Fig. ISO.— Elevation with Section of Parts. 



represents the profile of the stile around the panel s 
shows the line upon which, or the direction in which, 
the section is taken, said section being turned upon 
this line obliquely to the left. It is necessary to rule 
or line this section, the ruling being kept close to and 
inside the outline or profile. By placing the ruling in- 
side the profile no doubt can exist as to which parts 
are raised and which are depressed, for if at D the 
ruling were upon the other side of .the line from that 
shown the section D would indicate a depressed panel 
instead of a raised one. 

In the solution of the class of problems treated 
in Chapter VI, Section 1 (Miter Cutting), confusion 
often arises in the mind of the pattern cutter as to the 
proper position of a profile or of a miter line, which 



Linear Drawing. 



35 



confusion could never occur if all the necessary views 
were first drawn in accordance with the principles 
which this chapter is written to explain. A pro- 
file is always a section, and a miter line is either a part 
of an elevation projected from the section or part of 
another section bearing certain relations of hight or 
breadth to the first. A pattern is likewise always pro- 
jected — that is, carried off by right lines — from an 
elevation or plan the same as an elevation is projected 
from a section. 

It should also be remembered in this connection 
that the operation of developing a pattern is not com- 
pleted until its entire outline is drawn. The line form- 
ing its termination at the end opposite the miter cut, 



although simply a straight line, is properly derived 
from the elevation or plan used, the same as all points 
and other lines of the pattern. 

Much trouble is experienced through lack of 
knowledge of the principles of Linear Drawing, which 
if thoroughly understood could never result in such 
mistakes as producing a face miter where a return 
was intended or using the piece of metal from the 
wrong side of the miter cut. 

Too much emphasis cannot be placed upon the im- 
portance of thoroughly understanding the subject treated 
in this chapter, as such a knowledge comprehends 
within itself an answer to the many questions continually 
arising in the course of the pattern draftsman's labors. 



♦ 



CHAPTER IV. 



Geometrical Problems. 



In presenting this chapter to the student no at- 
tempt has been made to give a complete list of geo- 
metrical problems, but all those have been selected 
which can be of any assistance to the pattern drafts- 
man, and especial attention has been given in their 
solution to those methods most adaptable to his wants. 
They are arranged as far as possible in logical order 
and are classified under various sub-heads in such a 
manner that the reader will have no difficulty in finding 
what he wishes by simply looking through the pages, 
the diagrams given with each being sufficient to indi- 
cate the nature of the problem and, as it were, form a 
sort of index. 

1. To Draw a Straight Line Parallel to a Given Line 
and at a Given Distance from it, Using: the Compasses and 
Straight-Edge— In Fig. 131, let C D be the given line 
parallel to which it is desired to draw another straight 
line. Take any two points, as A ami B, in the given 
line as centers, and, with a radius equal to the given 
distance, describe the arcs x x and y y. Draw a line 
touching these arcs, as shown at E F. Then E F will 
be parallel to C D. 

2. To Draw a Line Parallel to Another by the Use of 
Triangles or Set-Squares.— In Fig. 132, let A B be the 
line parallel to which it is desired to draw another. 
Place one side of a triangle or set-square, F 1 , against 
it, as indicated by the dotted lines. While holding 
F 1 firmly in this position, bring a second triangle, or any 
straightedge, E, against one of its other sides, as shown. 
Then, holding the second triangle firmly in place, slide 
the first away from the given line, keeping the edges 
of the two triangles in contact, as shown in the figure. 
Against the same edge of the first triangle that was 
placed against the given line draw a second line, as 
shown by C D. Then C D will be parallel to A B. 
In drawing parallel lines by this method it is found 
advantageous to place the longest edges of the tri- 
angles against each other, and to so place the two in- 
struments that the movement of one triangle against 



the other shall be in a direction oblique to the lines 
to be drawn, as greater accuracy is attainable in this 
way. 

3. To Erect a Perpendicular at a Given Point in a 
Straight Line by Means of the Compasses and Straight- 
Edge. — In Fig. 133, let A B represent the given 
straight line, at the point C in which it is required to 
erect a perpendicular. From C as a center with any 
convenient radius strike small arcs cutting A B, as 
shown by D and B. With D and B as centers, and 
with any radius longer than the distance from each of 
these points to C, strike arcs, as shown by x x and y y. 
From the point at which these arcs intersect, E, draw a 
line to the point C, as shown, Then E C will be per- 
pendicular to A B. 

4. To Erect a Perpendicular at or near the End of 
a Given Straight Line by Means of the Compasses and 
Straight-Edge.— First Method.— In Fig. 131, let A B be 
the given straight line, to which, at the point P, situ- 
ated near the end, it is required to erect a perpendic- 
ular. Take any point ((J) outside of the line A B. 
With C as center, and with a radius equal to the 
distance from C to P, strike the arc, as shown, cutting the 
given line A B in the point P, continuing it till it also 
cuts in another point, as at E. From E, through the 
center, C. draw the line E F, cutting the arc, as shown 
at F. Then from the point F, thus determined, draw 
a line to P, as shown. The line F P is perpendicular 
to A B. 

5. To Erect a Perpendicular at or near the End of 
a Given Straight Line by Means of the Compasses and 
Straight-Edge.— Second Method.— In Fig. 135, let B A 
be the given straight line, to which, at the point P, it is 
required to erect a perpendicular. From the point P, 
with a radius equal to three parts, by any scale, de- 
scribe an arc, as indicated by x x. From the same 
point, with a radius equal to four parts, cut the line 
B A in the point C. From the point C, with a radius 
equal to five parts, intersect the arc first drawn by the 



Geometrical Problems. 



37 



are y y. From the point of intersection D draw the line 
D P. Then D P will be perpendicular to B A. 

6. To Draw a Line Perpendicular to Another Line 
by the Use of Triangles or Set Squares.— In Fig. 136, 
let C D be the given line, perpendicular to which it is 
required to draw another line. Place one side of a 
triangle, B, against the given line, as shown. Bring 
another triangle, A, or any straight edge, against the 
long side or hypolhenuse of the triangle B, as shown. 
Then move the triangle B along the straight edge or 
triangle A, as indicated by the dotted lines, until the 
opposite side of B crosses the line C D at the required 



sides of the given line A B. A line drawn through 
these points of intersection, as shown by G H, will 
bisect the line A B, or, in other words, divide it into 
two equal parts. 

8. To Divide a Straight Line into Two Equal Parts 
by the Use of a Pair of Dividers.— In Fig. 138, it is re- 
quired to divide the line A B into two equal parts, or 
to find its middle point. Open the dividers to as near 
half of the given line as possible by the eye. Place 
one jDoint of the dividers on one end of the line, as at 
A. Bring the other point of the dividers to the line, 
as at C, and turn on this point, carrying the first 



E— rr= 



Fig. 181.— To Draw a Straight Line' Par- 
allel to a Given Straight Line, and at a 
Given Distance from it, Using the Com- 
passes and a Straight-Edge. 




Fig. 132. — To Draw a Line Parallel to 
Another by the Use of Triangles or 
Set Squares. 



\E/ 



D C 

Fig. 133. — To Erect a Perpendicular at 
a Given Point in a Straight Line, 
Using the Compasses and Straight- 
Edge. 



A F 



c/ 



/ 



i^ 



/p 



\ 



y 



Fig. 134 — To Erect a Perpendicu- 
lar at or near the End of a 
Given Straight Line, Using the 
Compasses and Straight-Edge. 
First Method. 



/ 



D/ 

y 



X 



*\ 






\ 



X 



N 




Fig. 135.— To Erect a Perpendicular 
at or near the End of a. Given 
Straight Line. Second Method. 



Fig. 136. — To Draw a Line Perpen- 
dicular to Another by the Use of 
Triangles. 



point. "When against it, draw the line E F, as shown. 
Then E F is perpendicular to C D. It is evident that 
this rule is adapted to drawing perpendiculars at any 
point in the given line, whether central or located near 
the end. Its use will be found especially convenient 
for erecting perpendiculars to lines which run oblique 
to the aides of the drawing board. 

7. To Divide a Given Straight Line into Two Equal 
Parts, with the Compasses, by Means of Arcs.— In Fig. 137, 
let it be required to divide the straight line A B into two 
equal parts. From the extremes A and B as centers, and 
with any radius greater than one-half of A B, describe 
the arcs dfand. a e, intersecting each other on opposite 



around to D. Should the point D coincide with the 
other end of the line, the division will be correct. 
But should the point D fall within (or without) the 
end of the line, divide this deficit (or excess) into two 
equal parts, as nearly as is possible by the eye, and 
extend (or contract) the opening of the dividers to this 
point and apply them again as at first. Thus, finding 
that the point D still falls within the end of the line, the 
first division is evidently too short. Therefore, divide 
the deficit D B by the eye, as shown by E, and in- 
crease the space of the dividers to the amount of one 
of D E. Then, commencing again at A, step off as 
before, and finding that upon turning the dividers 



38 



Tlie New Metal Worker Pattern Booh. 



upon the point F the other point coincides with the 
end of the line B, F is found to be the middle point in 
the line. Iu some cases it may be necessary to repeat 
this operation several times before the exact center is 
obtained. 

9. To Divide a Straight Line into Two Equal Parts 
by the Use of a Triangle or Set Square.— In Fig. 139, 
let A B be a given straight line. Place a T-square 





\ ( 


?/* 








/ 


\ 








/ 


\ 








/ 


\ 








/ 


\ 








1 


\ 








1 


\ 








1 
1 
i 
1 


\ 
\ 
\ 
i 






A 


1 

1 

| 






R 


H 


1 


1 




■■■- D 




I 
\ 
\ 
\ 
\ 
\ 
\ 
\ 
\ 


/ 
/ 
/ 
/ 

! 
t 
/ 
/ 
/ 
/ 








; 


\ 








e'\ 


rf 







Fig. 1S7. — To Divide a Straiqht Live into Two Equal Parts 
by Means of Arcs. 

or some straight edge parallel to A B. Then bring 
one of the right-angled sides of a set square against it, 
and slide it along until its long side, or hypothenuse, 
meets one end of the line, as A. Draw a line along 
the long side of the triangle indefinitely. Reverse the 
position of the set square, as shown by the dotted lines, 
bringing its long side against the end, B, of the given 




Fig. ISS.—To Bisect a Straight Line by the Use of the 
Dividers. 

straight line, and in like manner draw a line along its 
long side cutting the first line. Next slide the set 
square along until its vertical side meets the intersec- 
tion of the two lines, as shown at C, from which point 
drop a perpendicular to the line A B, cutting it at D„ 
Then D will be equidistant from the two extremities 
A and B. 

10. To Divide a Given Straight Line into Any Num- 
ber of Equal Parts.— In Fig. 140, let A B be a given 
straight, line to be divided into equal parts, in this case 



eight. From one extremity in this line, as at A, draw 
a line, as either A C or A D, oblique to A B. Set the 
dividers to any convenient space, and step off the 
oblique line, as A C, eight divisions, as shown by a b 
c d, etc. From the last of the points, A, thus obtained, 
draw a line to the end of the given line, as shown by 
A A 5 . Parallel to this line draw other lines, from each 
of the other points to the given line. The divisions 




Fig. lS9.—To Bisect a Straight Line by the Use of a Triangle. 

thus obtained, indicated in the engraving by a 3 V c% 
etc., will be the desired spaces in the given line. It is 
evident by this rule that it is immaterial, except as a 
matter of convenience, to what space the dividers are 
set. The object of the second oblique line in the en- 
graving is to illustrate this. Upon A C the dividers 
were set so as to produce spaces shorter than those re- 




Fig. HO. — To Divide a Given Straight Line into Any Number 
of Equal Parts. 

quired in the given line A B, while in A D the spaces 
were made longer than those required in the given line. 
By connecting the last point of either line with the 
point B, as shown by the lines h A 2 and A' A 3 , and draw- 
ing lines from the points in each line parallel to these 
lines respectively, it will be seen that the same divi- 
sions are obtained from either oblique line. 

11. To Divide a Straight Line Into Any Number of 
Equal Parts by Means of a Scale. — It may be more con- 
venient to transfer the length of a given line to a slip of 



Geometrical Problems. 



39 



paper, and by laving the paper across a scale, as shown 
in Fig. 141, mark the required dimensions upon it, 
and afterward transfer them to a given line, than to 
divide the line itself by one of the methods explained 
for that purpose. ' It may also occur that it is desirable 
to divide lines of different lengths into the same num- 
ber of equal parts, or the same lengths of lines into 
different numbers of equal parts. Such a scale as is 
shown in Fig. 141 is adapted to all of these purposes. 
The scale may be ruled upon a piece of paper or upon 
a sheet of metal, as is preferred. The lines may be all 
of one color, or two or more colors may be alternated, 
in order to facilitate counting the lines or following 
them by the eye across the sheet. In size, the scale 
should be adapted to the special purposes for which it 
is intended to be used. By the contrast of two colors 




Fig HI. — To Divide a Straight Line, into Any Number of 
Equal Parts by Means of a Scale. 

in ruling the lines, one scale may be adapted to both 
coarse and fine work. For instance, if the lines are 
ruled a quarter of an inch apart, in colors alternating 
red and blue, in fine work all the lines in a given space 
may be used, while in large work, in which the di- 
mensions are not required to be so small, either all the 
red or all the blue lines may be used, to the exclusion 
of those of the other color. Let it be required to di- 
vide the line A B in Fig. 141 into thirty equal parts. 
Transfer the length A B to one edge of a slip of paper, 
as shown by A 1 B : , and placing A 1 against the first line 
of the scale, carry B 1 to the thirtieth line. Then mark 
divisions upon the edge of the strip of paper opposite 
each of the several lines it crosses, as shown. Let it 
be required to divide the same length A B into fifteen 
equal parts by the scale. Transfer the length A B to 



a straight strip of paper, as before. Place A' against 
the first line and carry B 2 against the fifteenth line, 
as shown. Then mark divisions upon the edge of the 
paper opposite each line of the scale, as shown. 

12. To Divide a Given Angle into Two Equal Parts.— 
In Fig. 142, let A C B represent any angle which it is 




Fig. W —To Bisect a Given Angle. 

required to bisect. From the vertex, or point C, as 
center, with any convenient radius, strike the arc D E, 
cutting the two sides of the angle. From D and E as 
centers, with any radius greater than one-half the 
length of the arc D E, strike short arcs intersecting at 
G, as shown. Through the point of intersection, G, 
draw a line to the vertex of the angle, as shown by 
F C. Then F C will divide the angle into two equal 
parts. 

13. To Trisect an Angle— No strictly geometrical 
method of solving this problem has ever been discov- 
ered. The following method, partly geometrical and 
partly mechanical, is, however, perfectly accurate and 
can be used to advantage whenever it becomes neces- 
sary to find an exact one-third or two-thirds of an 
angle : 

Let ABC, Fig. 143, be the angle of which it is 




Fig. 143. — To Trisect a Given Angle. 

required to find one-third. Extend one of its sides 
beyond the vertex indefinitely, as shown by B E, and 
upon this line from B as center with any convenient 
radius describe a semicircle A C D, cutting both sides 
of the angle. Place a straight edge firmly against the 



40 



Tlie New Metal Worker Pattern Book. 



extended side as at F, and a pin at the point C. On 
another straight edge (G) having a perfect corner at E, 
set off from one end a distance equal to the radius 
of the semicircle as shown by point x\ and placing 
this straight edge, with the end upon which the radius 
was set off, against the other straight edge (F) and its 
edge near the other end, against the pin at the point C, 
all as shown, slide it along until the mark x comes to 
the semicircle establishing the point D. Draw the line 
D B, then the angle DEB will be one-third of the 
angle ABC, and C D B will be two-thirds of it. 

14. To Find the Center from which a Given Arc 
is Struck.— In Fig. 144, let A B C represent the given 





Fig. HA.— To Find the Center 
from which a Given Arc is 
Struck. 



Fig. 140— The Chord and Eight 
of a Segment of a Circle Being 
Given, to find the Center from 
which the Arc may be Struck. 



arc, the center from which it was struck being un- 
known and to be found. From any point near the 
middle of the arc, as B, with any convenient radius, 
strike the are F G, as shown. Then from the points 
A and C, with the same radius, strike the intersecting 
arcs I H and E D. Through the points of intersec- 
tion draw the lines K M and L M, which will meet in 
M. Then M is the center from which the given arc 
was struck. Instead of the points A and C being 
taken at the extremities of the arc, which would be 
quite inconvenient in the case of a long arc, these 
points may be located in any part of the arc which is 
most convenient. The greater the distance between 
A and B and B and C, the greater will be the accu- 
racy of succeeding operations. The essential feature 
of this rule is to strike an arc from the middle one of 
the points, and then strike intersecting arcs from the 
other two points, using the same radius. It is not 
necessary that the distance from A to B and from B 
to C shall be exactly the same. 



15. To Find the Center from which a Given Arc is 
Struck by the Use of the Square.— In Fig. 145, let A 
B C be the given arc. Establish the point B at pleas- 
ure and draw two chords, as shown by A B and B C. 
Bisect these chords, obtaining the points E and D. 
Place the square against the chord B C, as shown in 
the engraving, bringing the heel against the middle 
point, D, and scribe along the blade indefinitely. 
Then place the square, as shown by the dotted lines, 
with the heel against the middle point, E, of the 
second chord, and in like manner scribe along the 
blade, cutting the first line in the point F. Then F 
will be the center of the circle, of which the are ABC 




Fig. 145.— To Find the Center from which a Given Arc is 
Struck, by the Use of the Square. 



is a part. This rule will be found very convenient for 
use in all cases where the radius is less than 24 inches 
in length. 

16. The Chord and Hight of a Segment of a Circle 
being Given, to Find the Center from which the Arc 
may be Struck.— In Fig. 146, let A B represent the 
chord of a segment or arc of a circle, and D C the rise 
or hight. It is required to find a center from which 
an arc, if struck, will pass through the three points 
A, D and B. Draw A D and B D. Bisect A D, as 
shown, and prolong the line H L indefinitely. Bisect 
D B and prolong I M until it cuts H L, produced in 
the point E. Then E, the point of intersection, will 
be the center sought. It will be observed that by 
producing D C, and intersecting it by either H L or 



Geometrical Problems 



41 



I M prolonged, the same point is found. Therefore, 
if preferred, the bisecting of either A D or D B may 
be dispensed with. A practical application of this 
rule occurs quite frequently in cornice work, in the 
construction of window caps and other similar forms, 
to fit frames already made. In the conveying of or- 
ders from the master builder or carpenter to the cor- 
nice worker, it is quite customary to describe the shape 
of the head of the frames which the caps are to fit by 
stating that the width is, for example, 36 inches, and 
that the rise is 4 inches. To draw the shape thus de- 
scribed, proceed as follows : Set off A B equal to 36 
inches, from the center of which erect a perpendicular, 
D C, which make equal to 4 inches. Continue D C in 
the direction of E indefinitely. Draw A D, which 
bisect, as shown, and draw H L, producing it until it 
cuts D C prolonged, in the point E. Then with E as 
center and E D as radius, strike the arc ADB. 

if. To Strike an Arc of a Circle by a Triangular 
Guide, the Chord and Hight Being Given.— In Fig. 147, let 




Fig. llfl. — To Strike an Arc of a Circle by a Triangular 
Chiide. 



A D be the given chord and B F the given bight. The 
first step is to determine the shape and size of the tri- 
angular guide. Connect A and F, as shown. From 
F, parallel to the given chord A D, draw F G, making 
it in length equal to A F, or longer. Then A F Gr, as 
shown in the engraving, is the angle of the triangular 
guide to be used. Construct the guide of any suitable 
material, making the angle of two of its sides equal to 
the angle AFC Drive pins at the points A, F and 
D. Place the guide as shown. Put a pencil at the 
point F. Shift the guide in such a manner that the 
pencil will move toward A, keeping the guide at all 
times against the pins A and F. Then reversing, shift 
the guide so that the pencil at the point F will move 
toward D, keeping the guide during this operation 
against the pins F and D. By this means the pencil 
will be made to describe the arc A F D. It may be 
interesting to know that if the angle F of the triangu- 
lar guide be made a right angle, the arc described by 
it will be a semicircle. By these means, then, a steel 
square may be used in drawing circles, as illustrated 
in Fig. 148, the pins being placed at A, B and C. 



18. To Draw a Circle Through any Three Given 
Points not in a Straight Line.— Iu Fig. 149, let A, D and 

E be any three given points not in a straight line, 
through which it is required to draw a circle. Con- 
nect the given points by drawing the lines A D and D 
E. Bisect the line A D by F C, drawn perpendicular 
to it, as shown. Also bisect D E by the line G C, as 
shown. Then the point C, at which these lines meet, 
is the center of the required circle. 

19. To Erect a Perpendicular to an Arc of a Circle, 
without having Recourse to the Center.— In Fig. 150, let 
A D B be the arc of a circle to which it is required to 
erect a perpendicular. With A as center, and with 
any radius greater than half the length of the given 
arc, describe the arc x x, and with B as center, and 
with the same radius, describe the arc y y, intersecting 




Fig. 14S. — To Describe a Semicircle with a Steel Square, 

the arc first struck, as shown. Through the points of 
intersection draw the line F E. Then F E will be 
perpendicular to the arc, and if sufficiently produced 
will reach the center from which the arc A B is drawn. 

20. To Draw a Tangent to a Circle or Arc of a 
Circle at a given Point without having Recourse to the 
Center.— In Fig. 151, let A D B be the arc of a circle, 
to which a tangent is to be drawn at the point D.- 
AVith D as center, and with any convenient radius, 
describe the arc A F B, cutting the given arc in the 
points A and B. Join the points A and B, as shown. 
Through D draw a straight line parallel to A B, as 
shown by E H, then E II will be the required tangent. 

21. To Ascertain the Circumference of a Given Circle. 
— In Fig. 152, let A D B C be the circle, equal to the 
circumference of which it is required to draw a straight 
line. Draw any two diameters at right angles, as 
shown by A B and D C. Divide one of the four arcs, 
as, for instance, D B, into eleven equal parts, as shown. 



42 



The New Metal Worker Pattern Book. 



From 9, the second of these divisions from the point 
B, let fall a perpendicular to A B, as shown by 9 F. 
To three times the diameter of the circle (A B or D C) 
add the length 9 F, and the result will be a very close 
approximation to the length of the circumference. This 
rule, upon a diameter of 1 foot, gives a length of about 
y 3 ¥ ths of an inch in excess of the actual length of the 
circumference. 

22. To Draw a Straight Line Equal in Length to the 
Circumference of any Circle or of any Part of a Circle- 



dividers are placed upon the line, no perceptible curve 
shall exist between them, and, beginning at one end 
of the curve, step to the other end of the same, or so 
near the end that the remaining space shall be less than 
that between the points of the dividers, then beginning 
at the end of any straight line step off upon it the same 
number of spaces, after which add to them the remain- 
ing small space of the curve by measurement with the 
dividers. This will be found the quickest and most 
accurate of any method for the pattern cutters' use. 




Fig. 149. — To Draw a Circle Through 
any Three Given Points Not in a 
Straight Line 





Fig. 150.— To Erect a Perpendicular to 
an Arc of a Circle. 



Fig. 151.— To Draw a Tangent to a Circle 
or Arc. 






Fig. 152. — To Ascertain the Circumfer- 
ence of a Given Circle. 



Fig. 15S. — To Inscribe an Equilateral 
Triangle within a Given Circle. 



Fig. 154. — To Inscribe a Square within 
a Given Circle. 



Various approximate rules, similar to the one given in 
the problem above, for performing these operations are 
known and sometimes used among workmen, but can- 
not be recommended here because in using them con- 
siderable time and trouble is required to obtain a result 
which is not accurate when obtained, thus rendering 
such methods impracticable. The simplest and most 
accurate method for obtaining the length of any curved 
line is as follows : Take between the points of the 
dividers a space so small that when the points of the 



The most common rules in use for the construc- 
tion of polygons, whether drawn within circles or 
erected upon given sides, are those which employ the 
straight-edge and compasses only. Other instruments 
may also be employed to great advantage, as will be 
shown further on, leaving the student to decide which 
method is the most suited to any case he may have in 
hand. Accordingly, the construction of polygons will 
be treated under three different heads arranged accord- 
ing to the tools employed. 



Geometrical Problems. 

THE CONSTRUCTION OF REGULAR POLYGONS. 



43 



I.— BY THE USE OF COMPASSES AND STRAIGHT-EDGE. 

23. To Inscribe an Equilateral Triangle within a 
Given Circle.— In Fig. 153, let ABD be any given 
circle within which an equilateral triangle is to be 
drawn. From any point in the circumference, as E, 
with a radius equal to the radius of the circle, describe 
the arc D C B, cutting the given circle in the points 
D and B. Draw the line D B, which will be one side 
of the required triangle. From D or B as center, and 
with D B as radius, cut the circumference of the given 
circle, as shown at A. Draw A B and A D, which 
will complete the figure. 

24. To Inscribe a Square within a Given Circle. — 
In Fig. 154, let A G B D be any given circle within 
which it is required to draw a square. Draw any two 



Circle.— In Fig. 156, let A B D E F G be any given 
circle within which a hexagon is to be drawn. From 
any point in the circumference of the circle, as at A, 
with a radius equal to the radius of the circle, de- 
scribe the arc C B, cutting the circumference of the 
circle in the point B. Connect the points A and B. 
Then A B will be one side of the hexagon. With the 
dividers set to the distance A B, step off in the cir- 
cumference of the circle the points G, F, E and D. 
Draw the connecting lines A Gr, Gr F, F E, E D and 
D B, thus completing the figure. By inspection of 
this figure it will be noticed that the radius of a circle 
is equal to one side of the regular hexagon which may 
be inscribed within it. Therefore set the dividers to 
the radius of a circle and step around the circumfer- 
ence, connecting the points thus obtained. 






Fig. 155 — To Inscribe a Regular Penta- 
gon within a Given Circle. 



Fig. 156. — To Inscribe a Regular Hexa- 
gon within a Given Circle. 



Fig. 157. — To Inscribe a Regular Hepta- 
gon within a Given Circle. 



diameters at right angles with each other, as C D and 
A B. Join the points C B, B D, D A and A C, which 
will complete the required figure. 

25. To Inscribe a Regular Pentagon within a Given 
Circle. — In Fig. 155, A D B C represents a circle in 
which it is required to draw a regular pentagon. Draw 
any two diameters at right angles to each other, as 
A B and DC. Bisect the radius A H, as shown at E. 
With E as center and E D as radius strike the arc D 
F, and with the chord D F as radius, from D as center, 
strike the arc F Gr, cutting the circumference of the 
given circle at the point G. Draw D G, which will 
equal one side of the required figure. With the 
dividers set equal to D G, step off the spaces in the 
circumference of the circle, as shown by the points 
I K and L. Draw D I, I K, K L and L G, thus com- 
pleting the figure. ' 

26. To Inscribe a Regular Hexagon within a Given 



27. To Inscribe a Regular Heptagon within a Given 

Circle.— In Fig. 157, let F A G B H I K L D be the 
given circle. From any point, A, in the circumfer- 
ence, with a radius equal to the radius of the circle, 
describe the arc BCD, cutting the circumference of 
the circle in the points B and D. Draw the chord B D. 
Bisect the chord B D, as shown at E. With D as 
center, and with D E as radius, strike the arc E F, 
cutting the circumference in the point F. Draw D F, 
which will be approximately one side of the heptagon. 
With the dividers set to the distance D F, set off in 
the circumference of the circle the points G, H, I, K 
and L, and draw the connecting lines, F G, G H, H I, 
IK, K L and L D, thus completing the figure. 

28. To Inscribe a Regular Octagon within a Given 
Circle.— In Fig. 15 S, let B ID F A G E H be the 
given circle within which an octagon is to be drawn. 
Draw any two diameters at right angles to each other, 



44: 



Tlie New Metal Worker Pattern Book. 



as B A and D E. Draw the chords D A and A E. 
Bisect D A, as shown, and draw L H. Bisect A E 
and draw K I. Then connect the several points in the 
circumference thus obtained by drawing the lines D I, 
I B, B H, H E, E G, G A, A F and F D, which will 
complete the figure. 

29. To Inscribe a Regular Nonagon within a Given 
Circle. — In Fig. 159, let A H K M be the given circle. 
Through the center C, draw any diameter A K, which 
extend indefinitely in either direction. From A as 
center, with the radius of the given circle, describe an 
arc, cutting the circumference at B. The arc A B 
being one-sixth of the circumference, may now be tri- 
sected, as explained in Problem 13 of this chapter, viz. : 
Set off on any straight edge the distance E D equal to 
the radius of the circle, and bringing the straight edge 
against the point B, slide the point E along the diam- 



eter 



extended 




K, etc. Draw D H, H I, I. J, etc., completing the figure. 

31. To Inscribe a Regular Polygon of Fifteen Sides 

in a Given Circle.— In Fig. 161, let A P L B be the 

given circle. First draw any two radii, as C A and 
C B, at right angles to each other, and bisect one of 
them, as CB, at D, and draw D A. With D as center, 
and C D as radius, describe an arc, cutting D A at E, 
and with A as center, and A E as radius, describe an 
arc E F, cutting the circumference at F. Then will A 
F be one-tenth of the circumference of the given circle. 
Now with A as center, and A C as radius, describe an 
arc C G, cutting the circumference at G. Then, since 
A G is one-sixth, and A F one-tenth of the circum- 
ference, F G must be the difference, which is one- 
fifteenth. Therefore with the dividers set to F Gr, 
divide the circumference into fifteen equal spaces, as 
shown by the points H, I, J, etc., and complete the 
figure by drawing the lines F G, G H, H I, etc. 

D 



B-|- 




Fig, 



15S. — To Inscribe a Regular Octa- 
gon within a Given Circle. 



159. — To Inscribe a Regular Nona- 
gon within a Given Circle. 



Fig. 160. — To Inscribe a Regular Deca- 
gon within a Given Circle. 



circumference, all as shown by E D B, and draw E B. 
Now from C draw C F parallel to E B, cutting the cir- 
cumference at F. Then will B F be equal to two- 
thirds of B A or one-ninth of the circumference. Set 
the dividers to this distance and space off the circum- 
ference, obtaining the points G, H, I, etc. Draw the 
lines B F, F G, G H, etc., completing the figure. 

30. To Inscribe a Regular Decagon within a Given 
Circle.— In Fig. 160, let D A E B be any given circle 
in which a decagon is to be drawn. Draw any two 
diameters through the circle at right angles to each 
other, as shown by B A and D E. Bisect B C, as 
shown at F, and draw F D. From F as center, with 
F C as radius, describe an arc C G, cutting F D at G. 
From D as center, with D G as radius, describe the 
arc G H, cutting the circumference at the point H. 
Set the dividers to the distance D H and divide the 
circle into ten equal spaces, shown by the points I, J, 



32. To Inscribe a Regular Dodecagon within a Given 
Circle. — In Fig. 162, let M F A I be any given circle 
in which a dodecagon is to be drawn. From any 
point in the circumference, as A, with a radius equal 
to the radius of the circle, describe the arc C B, cut- 
ting the circumference in the point B. Draw the 
chord A B, which bisect as shown, and draw the line 
C, cutting the circumference in the point D. Draw 
A D, which will then be one side of the given figure. 
With the dividers set to this space step off in the cir- 
cumference the points B, I, N, H, M, G, L, F, K and 
E, and draw the several chords, as shown, thus com- 
pleting the figure. 

33. General Rule for Inscribing any Regular Polygon 
in a Given Circle. — Through the given circle draw any 
diameter. At right angles to this diameter draw a 
radius. Divide that radius into four equal parts, and 
prolong it outside the circle to a distance equal to three 



Geometrical Problems. 



45 



of those parts. Divide the diameter of the circle into 
the same number of equal parts as the polygon is to 
have sides. Then from the end of the radius prolonged, 
as above described, through the second division in the 
diameter, draw a line cutting the circumference. Con- 
nect this point in the circumference and the nearest 
end of the diameter. The line thus drawn will be one 
side of the required figure. Set the dividers to this 
space and step off on the circumference of the circle 



outside the circle to the extent of three of those parts, 
as shown by a b c, thus obtaining the point c. From 
c, through the second division in the diameter, draw 
the line c H, cutting the circumference in the point 
H. Connect H and E. Then H E will be one side 
of the required figure. Set the dividers to the dis- 
tance H E and step off the circumference, as shewn, 
thus obtaining the points for the other sides, and draw 
the connecting arcs, all as illustrated in the figure. 





Fig. 161 —To Inscribe a Regular Polygon 
of Fifteen Sides within a Given Circle. 



-— %C 



Fig. 162. — To Inscribe a Regular Dodec- 
agon within a Given Circle. 




Fig. I6S.—T0 Inscribe a Regular Undec- 
agon within a Given Circle by the 
General Rule. 






Fig. I64. — Upon a Given Side to 
Construct an Equilateral Tri- 
angle. 



Fig. 165. — To Construct a Triangle, the 
Length of the Three Sides being 
Given. 



Fig. 166 — Upon a Given Side to draw a 
Regular Pentagon. 



the remaining number of sides and draw connecting 
lines, which will complete the figure. 

34. To Inscribe a Regular Polygon of Eleven Sides 
(Undecagon) within a Given Circle by the General Rule. 

— Through the given circle, E D F G in Fig. 163, 
draw any diameter, as E F, which divide into the 
same number of equal parts as the figure is to have 
sides, as shown by the small figures. At right angles 
to the diameter just drawn draw the radius D K, which 
divide into four equal parts. Prolong the radius D K 



35. Upon a Given Side to Construct an Equilateral 

Triangle.— In Fig. 164, let A B represent the length 
of the given side. Draw any line, as C D, making it 
equal to A B. Take the length A B in the dividers, 
and placing one foot upon the point C, describe the 
arc E F. Then from D as center, with the same 
radius, describe the arc G BZ, intersecting the first arc 
in the point K. Draw K C and K D. Then C D K 
will be the required triangle. 

36. To Construct a Triangle, the Length of the Three 



46 



Tlie New Metal Worker Pattern Book. 



Sides being Given.— In Fig. 165, let A B, C D and E 

F be the given sides from which, it is required to con- 
struct a triangle. Draw any straight line, G H, mak- 
ing it in length equal to one of the sides, E F. Take 
the length of one of the other sides, as A B, in the 
compasses, and from one end of the line just drawn, 
as G, for center describe an arc, as indicated by L 
M. Then set the compasses to the length of third 
side, C D, and from the opposite end of the line 
first drawn, H, describe a second arc, as I K, intersect- 
ing the first iii the point 0. Connect G and O H. 
Then G H will be the required triangle. 

37. Upon a Given Side to Draw a Regular Pentagon. 
— In Fig. 166, let A B represent the given side upon 
which a regular pentagon is to be constructed. With 
B as center and B A as radius, draw the semicircle 



circumference of the circle, obtaining the points M 
and L. Draw A M, M L and L D, which will com- 
plete the figure. 

38. Upon a Given Side to Draw a Regular Hexagon. 

— In Fig. 167, let A B be the given side upon which 
a regular hexagon is to be erected. From A as center, 
and with A B as radius, describe the arc B C. From 
B as center, and with the same radius, describe the 
arc A C, intersecting the first arc in the point C. C 
will then be the center of the circle which will cir- 
cumscribe the required hexagon. "With C as center, 
and C B as radius, strike the circle, as shown. Set 
the dividers to the space A B and step off the circum- 
ference, as shown, obtaining the points E, G, F and D. 
Draw the chords A E, E G, G F, F D and D B, thus 
completing the required figure. 




Fig. 167. — Upon a Given Side to Draw 
a Regular Hexagon. 




A ^- S B 



Fig. 168. — Upon a Given Side to Draw 
a Regular Heptagon. 




Fig. 169. — Upon a Given Side to Draw 
a Regular Octagon. 



A D E. Produce A B to E. Bisect the given side 
A B, as shown at the point F, and erect a perpendic- 
ular, as shown by F C. Also erect a perpendicular 
at the point B, as shown by G H. With B as center, 
and F B as radius, strike the arc F G, cutting the per- 
pendicular H G iu the point G. Draw G E. With 
G as center, and G E as radius, strike the arc E H, 
cutting the perpendicular in the point H. With E as 
center, and E H as radius, strike the arc H D, cutting 
the semicircle A D E in the point D. Draw D B, 
which will be the second side of the pentagon. Bisect 
D B, as shown, at the point K, and erect a perpendic- 
ular, which produce until it intersects the perpendic- 
ular F C, erected upon the center of the given side in 
point F. Then C is the center of the circle which 
circumscribes the required pentagon. From C as cen- 
ter, and with C B as radius, strike the circle, as shown. 
Set the dividers to the distance A B and step off the 



39. Upon a Given Side to Draw a Regular Heptagon. 

— In Fig. 168, A B represents the given side upon 
which a regular heptagon is to be drawn. From B as 
center, and with B A as radius, strike the semicircle 
A E D. Produce A B to D. From A as center, and 
with A B as radius, strike the arc B F, cutting the 
semicircle in the point F. Through F draw F G per- 
pendicular to A B, which extend in the direction of C. 
From D as center, and with radius G F, cut the semi- 
circle in the point E. Draw the line E B, which is 
another side of the required heptagon. Bisect E B, and 
upon its middle point, H, erect a perpendicular, which 
produce until it meets the perpendicular erected upon 
the center of the given side A B, in the point C. Then 
C is the center of the circle which will circumscribe 
the required heptagon. From C as center, and with 
C B as radius, strike the circle. Set the dividers to 
the distance A B and step off the circumference, as 



Geometrical Problems. 



47 



shown, obtaining the points K, N, M and L. Draw 
the connecting arcs A K, K N, 1ST M, M L and L E, 
thus completing the figure approximately correctly. 
40, Upon a Given Side to Draw a Regular Octagon.— 

In Fig. 169, let A B represent the given side upon 
which a regular octagon is to be constructed. Produce 
A B indefinitely in the direction of D. From B as 
center, and with A B as radius, describe the semicircle 
A E D. At the point B erect a perpendicular to A B, 
as shown, cutting the circumference of the semicircle 
in the point E. Bisect the arc E D, obtaining the 
point F. Draw F B, which is another side of the re- 
quired octagon. Bisect the two sides now obtained 
and erect perpendiculars to their middle points, Gr and 
H, which produce until they intersect at the point C. 
C then is the center of the circle that will circumscribe 



thus obtaining the point G, and draw B Gr, which will 
be another side of the required figure. From the 
middle points, H and K, of the two sides now obtained, 
erect perpendiculars, which produce until they inter- 
sect at L. Then L is the center of a circle which 
will circumscribe the required nonagon. From L 
as center, with A L as radius, describe the circle 
A M Q. Set the dividers to the space A B and 
divide the circle in nine equal spaces, as shown by 
the points M, N, 0, etc., and draw the connecting 
lines GrM, MN, NO, OP, etc., thus completing the 
figure. 

42. Upon a Given Side to Draw a Regular Decagon. 
— In Fig. 171, let A B be the given side upon which 
a regular decagon is required to be drawn. Produce 
the line A B indefinitely in the direction of D, and 




L _ .k 





Fig. 170. — Upon a Given Side to Draw 
a Regular Nonagon. 



Fig. 171 — Upon a Given Side to Draw 
a Regular Decagon. 



l\ 
/I \ 
/ I \ 

' I \ 

-- F 
E 

Fig. 172.— Upon a Given Side, to Draw a 

Jiegulir Polygon having fifteen b ides. 



the octagon. From C as center, and with C B as radius, 
strike the circle, as shown. Set the dividers to the 
space A B and step off the circumference, obtaining 
the points L, K, M, and N". Draw the connecting 
arcs A L, L K, K M, M 0, O E" and 1ST F, thus com- 
pleting the required figure. 

41. Upon a Given Side to Draw a Regular Nonagon. — 
In Fig. 170, let A B be any given side upon which it 
is required to draw a regular nonagon. Produce A B 
indefinitely in the direction of D. Then from B as a 
center, and with A B as radius, draw the semicircle 



A C D, and from 



D as a center, and with the same 



radius, strike an arc, cutting the circumference at 0. 
Extend the line AB indefinitely in the direction of E, 
and trisect the arc C D, obtaining the point F. The 
method of doing this is all explained in Problems 13 
and 29 of this chapter. Bisect C F, or, what is the 
same thing, set off from D on D C the distance F C, 



from B as a center, and with B A as the radius, de- 
scribe the semicircle AHD. From its center point B 
erect the perpendicular line B H, which bisect in the 
point F, and draw F D. From F as a center, and with 
F B as radius, describe the arc B Gr, cutting F D at Gr, 
and from D as center, with D Gr as radius, describe 
the arc Gr E, cutting the semicircle at E, and draw B E. 
Then B E will be another side of the decagon. Upon 
the middle points P and Q of the two sides now 
obtained, erect perpendiculars, which produce until 
they intersect the point at C. Then C is the center 
from which a circle can be drawn which will circum- 
scribe the required decagon. From C as a center, 
and with C A as a radius, describe the circle, as shown. 
Set the dividers to the side A B and divide the circle, 
obtaining the points I, J, K, etc., as shown, and draw 



the lines E 
figure. 



I, I J, J K, etc., thus completing the 



48 



Tlie New Metal Worker Pattern Book. 



43. Upon a Given Side to Draw a Regular Polygon 
Having Fifteen Sides.— In Fig. 172, let A B be the given 
side on which it is desired to draw a regular polygon. 
Extend the line A B indefinitely in the direction of C, 
and from B as a center, and with A B as a radius, de- 
scribe the semicircle ADC. From C as a center, and 
with the same radius, describe an arc cutting the 
circumference in D and extend the arc indefinitely in 
the direction of E. Draw the line D C. Then from 
the point C draw C E at right angles to D C, 
cutting the arc just drawn at E. Bisect C E at F and 
draw F D. From F as center, with C F as radius, de- 
scribe an arc cutting F D at Gr. From D as center, 
with D G as radius, describe an arc cutting the semi- 
circle at H, and draw H B, which will be another side 
of the polygon. From the middle points, I and J of 
the two sides, erect lines perpendicular thereto, pro- 




a q"b d 

Fig. 173. — Upon a Given Side to Draw a Regular Dodecagon. 



ducing them till they intersect at K. Then will K be 
the center of a circle which will circumscribe the re- 
quired polygon. From K as center, with K A as 
radius, describe the circle A M P V. Set the dividers 
to the space A B, and divide the circle into equal 
spaces, as shown by the points L, M, N, etc., 
and draw the lines H L, L M, M N, etc., thus com- 
pleting the figure. 

44. Upon a Given Side to Draw a Regular Dodecagon. 
— In Fig. 173, let A B represent the given side upon 
which a regular dodecagon is to be drawn. Produce 
A B indefinitely in the direction of D. From B as 
center, and with B A as radius, describe the semicircle 
A F D. From D as center, and with D B as radius, 
describe the arc B F, cutting the semicircle in the 
point F. Draw F D, which bisect by the line V B, 
cutting the semicircle in the point E. Then E B is 
another side of the dodecagon. From the middle 



points of the two sides now obtained, as G and H, erect 
perpendiculars, as shown, cutting each other at the 
point C. This point of intersection, C, then is the 
center of the circle which will circumscribe the required 
dodecagon. From C as center, and with C B as radius, 
strike the circle, as shown. Set the dividers to the 
distance A B and space off the circumference, thus ob- 
taining the points L, P, M, S, 1ST, R, 0, K and I. 
Draw the connecting lines L P, P M, M S, S N, N R, 
R 0, OK, KI and I E, thus completing the figure. 

45. General Rule by which to Draw any Regular 
Polygon, the Length of a Side Being Given.— With a 
radius equal to the given side describe a semicircle, the 
circumference of which divide into as many equal parts 
as the figure is to have sides. From the center by 
which the semicircle was struck draw a line to the 




Fig. 174.— Upon a Given Side to Construct a Regular Polygon 
of TJiirteen Sides by the General Rule. 

second division in the circumference. This line will 
be one side of the required figure, and one-half of the 
diameter of the semicircle will be another, and the two 
will be in proper relationship to each other. There- 
fore, bisect each, and through their centers erect per- 
pendiculars, which produce until they intersect. The 
point of intersection will be the center of the circle 
which will circumscribe the polygon. Draw the circle, 
and setting the dividers to the length of one of the 
sides already found, step off the circumference, thus 
obtaining points by which to draw the remaining sides 
of the figure. 

46. To Construct a Regular Polygon of Thirteen Sides 
by the General Rule, the Length of a Side being Given.— 
In Fig. 174, let A B be the given side. With B as 
center, and with B A as radius, describe the semicircle 
A F G. Divide the circumference of the semicircle 



Geometrical Problems. 



49 



into thirteen equal parts, as shown by the small figures, 
1, 2, 3, 4, etc. From B draw a line to the second 
division in the circumference, as shown by B 2. Then 
A B and B 2 are two of the sides of the required figure, 
and are in correct relationship to each other. Bisect 
A B and B 2, as shown, and draw D C and E C through 
their central points, prolonging them until they inter- 
sect at the point C. Then C is the center of the circle 
which will circumscribe the required polygon. Strike 
the circle, as shown. Set the dividers to the space 
A B, and step off corresponding spaces in the circum- 
ference of the circle, as shown, and connect the several 
points so obtained by lines, thus completing the figure. 
47. Within a Given Square to Draw a Regular Octa- 
gon. — In Fig. 175, let ADBE be any given square 
within which it is required to draw an octagon. Draw 
the diagonals D E and A B, intersecting at the point 
C. From A, D, B and E as centers, and with radius 
equal to one-half of one of the diagonals, as A C, strike 
the several arcs H 1ST, G K, I M and L 0, cutting the 
sides of the square, as shown. Connect the points 
thus obtained in the sides of the square by drawing 
the lines G 0, II I, K L and M N, thus completing 
the figure. 

For general use a very convenient scale may be 
constructed, as shown in Fig. 176, from which half 
the length of one side of a polygon of any number of 
sides and of any diameter in inches and fractions of 










M 



A N~ E 

Fig. 175.— Within a Given Square to Draw a Regular Octagon. 

inches may readily be obtained. Draw the vertical 

line B and divide it into inches and parts of an inch. 

From these points of division draw horizontal lines ; 

from the point draw the following lines and at the 

following angles from the horizontal line P : 

A line at 75° for polygons having 12 sides. 

" 72° " " 10 " 

" 67*° " « 8 " 



A line at 60° for polygons having 6 sides. 
" 54° " "5 " 

" 45° " " 4 '< 

The figures on B will designate the radius of 
the inscribed circle measured from 0. The distance 
from B on any horizontal line to the oblique line de- 




Fig. 176 — Scale for Constructing Polygons of any Number 
of Sides, the Diameter of the Inscribed Circle Being 
Given in Inches. — Half Full Size, 



noting the required polygon will be half the length of 
a side of the polygon of the diameter indicated by the 
figure at the end of the horizontal line assumed. The 
distance from measured upon the oblique line to the 
assumed horizontal line will be the radius of the cir- 
cumscribed circle. 

In the engraving three polygons are drawn show- 
ing the application of the scale. 

II— BY THE USE OF THE T-SQUARE AND TRIAN- 
GLES OR SET-SQUARES. 

In the chapter upon terms and definitions under 
the word degree (clef. 68) and in some of those immedi- 
ately following the dimensions of the circle are de- 
scribed and their use explained; and in the chapter 
upon Drawing Tools and Materials (on page 21) the tri- 
angles or set-squares in common use are described and 
illustrated. As all regular polygons depend, for their 
construction, upon the equal division of the circle, 
some explanation of the application of the foregoing 
will serve to fix a few facts in the mind of the student 
and thus prepare him for the use of the set-square. 



50 



Hie New Metal Worker Pattern Book. 



A well-known and easily demonstrated geometrical 
principle is that the sum of the three interior angles of a 
triangle is equal to two right angles, or in other words, 
as a right angle is one of 90 degrees, if the three angles 
of any triangle be added together their sum will equal 
180 degrees. Hence, if one of the angles of a set- 
square be fixed at 90 degrees (which is done for con- 
venience in drawing perpendicular lines) the sum of 
the two remaining angles must also be 90 degrees, 
and, if then the two other angles be made equal, each 
will be 45 degrees, which is the half of 90 degrees. 
If, however, one of the other angles is fixed at 30 
(one-third of 90 degrees), the remaining angle must be 
60 degrees, as 30 -f- 60 = 90. 

By means, then, of the 45-degree and the 30 X 
60-degree triangles, the draftsman has at his command 



of the 45-degree triangle, as A E, is placed against the 
blade of the T-square, and the vertical division of the 
circle is drawn along the other short side C E. 

In Fig. ITS the vertical and horizontal divisions 
of the circle, A B and C D, are drawn as before, after 
which one of the shorter sides of the 45-degree triangle 
is placed against the T-square, and the long or oblique 
side E F is brought to the center of the circle and 
another division Gr I is drawn. By reversing the posi- 
tion of the triangle the last division H K is drawn, 
thus dividing the circle into eight equal parts. 

In Fig. 179, after drawing the divisions A B and 
C D as before, the 30 X 60-degree triangle is placed in 
the position shown at A E F, and the division E N is 
drawn along its hypothenuse or oblique side. By re- 
versing the position of the triangle, still keeping the 




Fig. 177. — Circle Divided into Four 
Equal Parts by the Use of a Triangle 
or Set-Square. 




Fig. 178. — Circle Divided into Eight 
Equal Parts by the Use of a 45-degree 
Triangle. 




Fig. 179. — Circle Divided into Twelve 
Equal Parts by the Use of a 80 x 60- 
degree Triangle. 



the means of drawing lines at angles of 90, 60, 45 and 
30 degrees, and by combination 75 degrees (45 -4- 30) 
and 15 degrees (90 — 75). With the 45-degree 
angle he can bisect a right angle, and with the 30 and 
60-degree angles he can trisect it. 

The pattern draftsman sometimes finds it con- 
venient to have a set-square in which the sharpest 
angle is one of 22-^ degrees (one-half of 45) for use in 
drawing the octagon in a certain position which will 
be referred to later. 

In Figs. 177, 178, 179 and 180 are illustrated 
the application of the foregoing, in which the circle is 
divided, by the use of the triangles above described, 
into four, eight and twslve equal parts. In Fig. 177 
the horizontal division A B of the circle is drawn by 
means of the T-square placed against the side of the 
drawing board, after which one of the shorter sides 



side A F against the blade of the T-square, the divi- 
sion J K may be drawn. Changing the position of the 
triangle now so that its shortest side comes against the 
blade of the T-square, as shown dotted at Gr H F, the 
division Gr M is drawn, and again reversing its posi- 
tion, still keeping its shortest side against the T-square, 
the last division I L may be drawn, thus dividing the 
circle into twelve equal parts. 

In Fig. ISO the circle is divided into eight equal 
parts, but differing from that shown above in this 
respect that, while in Fig. 178 two of the divisions lie 
parallel with the sides of the drawing board, in the 
latter case none of the divisions are parallel with the 
sides of the board or can be drawn with the T-square ; 
but if this method is used in drawing an octagon, as 
shown dotted in Fig. 180, four of the sides of the oc- 
tagon can be drawn with the T-square and the other 



(v-diin triad Problems. 



51 



four with the 45-degree triangle. The position of the 
22-J- X 67-J-degree triangle in drawing the divisions of 
the circle is shown at A B C and DEC. while the posi- 
tion of the 45-degree triangle in drawing the oblique 
sides of an octagon figure is shown at F. It will thus 
be seen that the 22^ X 67^-degree triangle is available 
in drawing accurately the miter line for all octagon 
miters. 

As a triangle in whatever form it may be con- 
structed is intended to be used by sliding it against 
the blade of the T-square, all the angles above men- 
tioned are calculated with reference to the lines drawn 
by the T-square. In practical use it will be found in- 
convenient in drawing such lines to actually bring the 
point of a set-square to the center of a circle. A 
better method, and one which makes use of the same 
principles, is shown in Fig. 181. The blade of the 




Fig. ISO. — Circle Divided into Eight Equal Parts by the 
Use of a 22% x 67%-degree Triangle. 

T-square is placed tangent to or near the circle, as shown 
by A B. One side of a 45-degree triangle is placed 
against it, as shown, its side C F being brought against 
the center. The line C F is then drawn. By reversing 
the triangle, as shown by the dotted lines, the line E 
D is drawn at right angles to C F, thus dividing the 
circle into quarters. 

A similar use of the 30 X 60-degree triangle is 
shown in Fig. 182, by which a circle is divided into 
six equal parts. Bring the blade of the T-square 
tangent to or near the circle, as shown by A B. Then 
place the set-square as shown by G B M, bringing the 
side G- B against the center of the circle, drawing the 
line D L. Then place it as shown by the dotted lines, 
bringing the side A H against the center, scribing the 
line F E. Then, by reversing the set-square, placing 
the side G M against the straight-edge, erect the per- 



pendicular C I, completing the division. The follow- 
ing are a few of the problems to which these principles 
may be advantageously applied. 

48. To Inscribe an Equilateral Triangle within a Given 

Circle.— In Fig. 183, let D be the center of the given 
circle. Set the side E F of a 30-degree set-square 
T-square, as shown, and move it 



against the 



along 




D F 

Fig. lSt— Proper Method of Using a i5-degree Triangle. 

until the side E G touches D. Mark the point B upon 
the circumference of the circle. Beverse the set-square 
so that the point E will come to the right of the side 
F G and move it along in the reversed position until 
the side E G again meets the point D, and mark the 
point C. Now move the T-square upward until it 
touches the point D, and mark the point A. Then 
A B and C are points which divide the circle into three 
equal parts. The triangle may be easily completed 
from this stage by drawing lines connecting A B, B C 
and C A, with any straight-edge or rule, but greater 
accuracy is obtained by the further use of the set- 
square, as follows : Place the side F G of the set- 
square against the T-square, as shown in Fig. 184, 




Fig. 182. — Method of Using a 30 x 60-degree Triangle in 
Dividing the Circle. 

and move it along until the side E G touches the points 
A and C, as shown. Draw A C, which will be one 
side of the required triangle. Set the side E F of the 
set-square against the T-square, and move it along until 
the side F G coincides with the points C and B. Then 
draw C B, which will be the second side of the trianerle. 



52 



The New Metal Worker Pattern Book. 



Place the side F G of the set-square against the T- 
square, with the side E F to the right, and move it 
along until the side E G coincides with the points A 
and B. Then draw A B, thus completing the figure. 
The same results may be accomplished with less work 
by first establishing the point A by bringing the 
T-square against the center, and then using the set- 
square, as shown in Fig. 184. The different methods 
are here given in order to more clearly illustrate the 
use of the tools employed. 




Fig. 183. 




Fig. 184. 

To Inscribe an Equilateral Triangle within a 
Given Circle. 



49. To Inscribe a Square within a Given Circle — 

Let D, in Fig. 185, be the center of the given circle. 
Place the side E F of a 45-degree set-square against 
the T-square, as shown, and move it along until the 
side E G meets the point D. Mark the points A and 
B. Reverse the set-square, and in a similar manner 
mark the points C and H. The points A, H, B and 
C are corners of the required square. Move the T- 
square upward until it coincides with the points A and 
H and draw A H, as shown in Fig. 1S6. In like man- 
ner draw C B. With the side E F of the set-square 
against the T-square, move it along until the side G F 



coincides with the points B and H, and draw B H. In 

a similar manner draw C A, thus completing the figure. 

50. To Inscribe a Hexagon within a Given Circle. — 

In Fig. 187, let be the center of the given circle. 
Place the side E F of a 30-degree set-square against 
the T-sq uare ) as shown. Move the set-square along 
until the side E G meets the point 0. Mark the points 
A and B. Reverse the set-square, and in like manner 
mark the points C and D. With the side F G of the 
set-square against the T- s q uare > move it along until 




Fig. 185. 




Fig. 186. 
To Inscribe a Square uithia a Given Circle. 

the side E F meets the point 0, and mark I and H. 
Then A, H, D, B, I and C represent the angles of the 
proposed hexagon. From this stage the figure may be 
readily finished by drawing the sides by means of these 
points, using a simple straight-edge; but greater ac- 
curacy is attained in completing the figure by the 
further use of the set-square, as shown in Fig. 188. 
With the side E F of the set-square against the 
T-square, as shown, draw the line H D, and by mov- 
ing the T-square upward draw the side C I. Reversing 
the set-square so that the point F is to the left of the 
point E, draw the side A H, and also, by shifting the 



Geometrical Problems. 



53 



T-square, the side I B. "With the edge E F of the 
set-square against the T-square, move it up until the 
side G F coincides with the points B and D, and draw 
the side B D. In like manner draw A C, thus com- 
pleting the figure. In this figure, as with the triangle, 
the same results may be reached by establishing the 
points H and I, by means of a diameter drawn at right 
angles to the T-square, as shown in the engravings, 
and, using it as a base, employing the set-square, as 
shown in Fig. 188. The first method shown is, how- 




Fig. 187. 




Fig. 188. 
To Inscribe a Regular Hexagon within a Given Circle. 

ever, to be preferred in many instances, on account of 
its great accuracy. 

51. To Inscribe an Octagon within a Given Circle- 
In Fig. 189, let K be the center of the given circle. 
Place a 45-degree set -square as shown in the engraving, 
bringing its long side in contact with the center, and 
mark the points E and A. Keeping it in the same po- 
sition, move it along until its vertical side is in contact 
with K and mark the points D and H. Beverse the 
set-square from the position shown in the engraving, 
and mark the points G and G. Move the T-square up- 
ward until it touches the point K, and mark the points 



B and F. Then A, H, G, F, E, D, C and B are cor- 
ners of the octagon. The figure may now be readily 
completed by drawing the sides, by means of these 
points, using any rule or straight-edge for the purpose, 




Fig. 189. — To Inscribe a Regular Octagon within a 
Given Circle. 



all as shown by A H, H G, G F, F E, E D, D C, C B 
and B A, or by means of a 22-J- X 6T£-degree set- 
square. 

52. To Draw an Equilateral Triangle upon a Given 

Side.— In Fig. 190, let A B be the given side. First 
bring the line A B at right angles to the blade of the 
T-square. Then set the edge C B of a 30-degree set- 
square against the T-square, and move it along until 
the edge B D meets the point B, and draw the line 
B F. Beverse the set-square, still keeping the side 
G B against the T-square, and move it along until the 




Fig. 190.— To Draw an Equilateral Triangle upon a 
Given Side. 



side B D meets the point A, and draw the line A F, 
thus completing the figure. 

53. To Draw a Square upon a Given Side.— In Fi°- 
191, let A B be the given side drawn at right angles 



54 



TJie New Metal Worker Pattern Booh. 



to the blade of the T-square. Set the edge E F of a 
45-degree set-square against the T-square, as shown, 
and move it along until the side E G meets the point 
B, and draw B I indefinitely. Beverse the set-square, 
and, bringing the side E G against the point A, draw A 




Fig. 191. — To Draiv a Square upon a Given Side. 



F indefinitely. Bring the T-square against the point 
B and draw B F, producing it until it meets the line 
A F in the point F. In like manner draw A I, meet- 
ing the line B I in the point I. Then, with the set- 
square placed as shown in the engraving, connect I 
and F, thus completing the required figure. 

54. To Draw a Regular Hexagon upon a Given Side. 
—In Fig. 192, let A B be the given side in a vertical 
position. Set the edge G H of a 30-degree set-square 
against the T-square, as shown, and move it along 




Fig. 192. — To Draw a Regular hexagon upon a Given Side. 

until the edge I G coincides with the point A, and 
draw the line A D indefinitely. Reverse the set- 
square, still keeping the edge G H against the 
T-square, and move it along until the side I G coin- 
cides with the point B, and draw B E indefinitely. 
These lines will intersect in the point O, which will be 



the center of the required figure. Still keeping the 
edge G H of the set-square against the T-square, move 
it along until the perpendicular edge I H meets the 
point 0, and through draw F C indefinitely. With 
the set-square in the position shown in the engraving 
slide it along until the edge I G meets the point 
B, and draw B C, producing it until it meets the line 
F C in the point C. Reverse the set-square, still 
keeping the edge G H against the T-square, and draw 
the line C D, producing it until it meets the line A D 
in the point D. Slide the set-square along until the 
side I H meets the point D, and draw the line B E, 
meeting the line B E in the point E. Move the set- 
square along until the edge I G meets the point A, and 
draw the line A F, meeting the line C F in the point 
F. Now bring the set-square to its first position and 




Fig. 193.— To Draw a Regular Octagon upon a Given Side. 

slide it along until the edge I G meets the points F and 
E, and draw F E, thus completing the required figure. 
55. To Draw a Regular Octagon upon a Given Side- 
In Fig. 193, let C D be the given side, drawn perpen- 
dicular to the blade of the T-square. Place one of the 
short sides of a 45-degree set-square against the 
T-square, as shown in the engraving. Move the set- 
square along until its long side coincides with the 
point C. Draw the line C B, and make it in length 
equal to C D. With the T-square draw the line A B, 
also in length eqiial to C D. Reverse the set-square, 
and bring the edge against the point A. Draw A H 
in length the same as C D. Still keeping a short side 
of the set-square against the T-square, slide it along 
until the other short side meets the point H, and draw 
H G, also of the same length. Then, using the long 
side of the set-square, draw G F of corresponding 
length. By means of the T-square draw F E, and by 
.reversing the set-square draw E D, both in length 



Geometrical Problems. 



55 



equal to the original side, C D, joining it in the point 
D, thus completing the required octagon. 

56. To Draw an Equilateral Triangle about a Given 
Circle —In Fig. 1 94, let O be the center of the given 
circle. Place the edge E F of a 30-degree set-square 
against the T-square, as shown, and move it along un- 
til the edge F G meets the center 0, and mark the 
point A upon the circumference of the circle. Reverse 
the set-square, still keeping the edge E F against the 
T-square, and in like manner mark the point B. Move 




Fig. lit I. 




Fig. 195. 
To Draw an Equilateral Triangle about a Given Circle. 

the T-square upward until it meets the point 0, and 
mark the point C. The required figure will be de- 
scribed by drawing lines tangent to the circle at the 
points A, B and C, which may be done in the manner 
following, as indicated in Fig. 195 Place the edge 
E G- of the set-square against the T-square, and slide it, 
along until the edge F Gr touches the circle in the point 
B. Draw I K indefinitely. Reverse the set-square, 
keeping the same edge against the T-square, and move 
it along until its edge F G touches the circle in the 
point A, and draw I L, intersecting I K in the point I, 



the other end being indefinite. Then, placing the edge 
F E of the set-square against the T-square, bring its 
edge E Gr against the circle in the point C, and draw 
L K, intersecting I D in the point L and I K in the 
point K, thus completing the figure. The first part of 
this operation is not really necessary. The sides of the 
set-square simply can be brought tangent to the circle, 
as in Fig. 195. 

57. To Draw a Hexagon about a Given Circle.— In 
Fig. 196, let be the center of the given circle. Place 




Fig. 196. 




Fig. 197. 
To Draw a Hexagon about a Given Circle. 

the edge E F of a 30-degree set-square against the 
J-square, and slide it along until the edge F Gr meets the 
point 0, and mark the points B and A. Reverse the set- 
square, still keeping the edge E F against the T-square, 
and in like manner mark the points C and D. Bring 
the edge of the T-square against O, and mark the points 
I and K. Then C, A, K, D, B and I are six points in 
the circumference of the circle, corresponding to the 
six sides of the required figure. The hexagon is com- 
pleted by drawing a side tangent to the circle at each 
of these several points, which may be done by using 



56 



The New Metal Worker Pattern Booh. 



the set-square as follows, and as shown in Fig. 197. 
"With the edge E G of the set-square against the 
T-square, bring the edge F G against the circle at the 
point C, as shown, and draw L M indefinitely. Re- 
verse the set-square, and in like manner bring it against 
the circle at the point A, and draw M N, cutting L M 
in the point M, and extending indefinitely in the direc- 
tion of N. Slide the set-square along until the edge 
E F meets the circle in the point K, and draw N P, 
intersecting M N in the point 1ST, and extending in the 




Fig. 198. 




Fig. 199. 
To Draw an Octagon about a Given Circle. 

direction of P indefinitely. With the set-square in 
its first position slide it along until the edge F G 
meets the circle in the point D, and draw R P, cut- 
ting N P in the point P, but being indefinite in 
the direction of E. Reverse the set-square, and in 
like manner draw R S tangent to the circle in the 
point B, cutting P R in the point R, and extending in 
the direction of S indefinitely. Slide the set-square 
along until its edge E F meets the circle in the point 
I, and draw S L, cutting R S in the point S and L M 
in the point L, thus completing the required figure. 



In this problem, as in the previous one, if care be taken 
the first part of the operation can be dispensed with by 
simply placing the triangle in proper position and draw- 
ing the sides of the figure tangent to the circle, as 
shown in Fig. 197. 

58. To Draw an Octagon about a Given Circle. — In 
Fig. 19S, let be the center of the given circle. With 
the edge E F of a 45-degree set-square against the T- 
square, as shown, move it along until the side E G 
meets the point 0, and mark the points A and B. 
Reverse the set-square, and in like manner mark the 
points C and D. Slide the set-square along until the 
vertical side G F meets the point 0, and mark the 
points H and I. Move the T-square up until it meets 
the point 0, and mark the points K and L. Then A, 
I, D, L, B, H, C and K are points in the circumfer- 
ence of the given circle corresponding to the sides of 




Fin. 2V0. — to Draw a Square about a Given Circle. 

the required figure. The octagon is then to be com- 
pleted by drawing lines tangent to the circle at these 
several points, as shown in Fig. 199, which may be 
done by the use of the set-square, as follows : With 
the edge E F of the set-square against the T-square, 
as shown, bring the edge E Gr against the circle in the 
point D, and draw M N" indefinitely. Sliding the set- 
square along until the vertical edge F Gr meets the 
circle in the point L, draw 1ST P, cutting M N in the 
point N, and extending in the opposite direction in- 
definitely. Reverse the set-square, and bringing the 
edge E G against the circle in the point B, draw P R, 
cutting N P in the point P, and extending indefinitely 
in the direction of R. Move the T-square upward un- 
til it meets the circle in the point H, and draw the line 
S R, meeting P R in the point R, and extending in- 
definitely in the opposite direction. Then, with the 
set-square placed as shown in the engraving, move it 



Geometrical Problems. 



57 



until its edge E G meets the circle in the point C, and 
draw S T, meeting S R in the point S, and continuing 
indefinitely in the direction of T. With the set-square 
in the same position, move it along until its edge G F 
meets the circle in the point K, and draw T IT, cutting 
S T in the point T, and extending in the opposite di- 
rection indefinitely. Reverse the set-square, and bring- 
ing its long side against the circle in the point A, draw 
U V, cutting T U in the point U, and continuing in- 
definitely in the opposite direction. Bring the J- 
souare against the circle in the point I, and draw V M, 
connecting U V and M N in the points V and M re- 
spectively, thus completing the figure. The above rule 
will be found very convenient for use, although, as 
the student may discover, the first part of the opera- 
tion is not absolutely necessary. 

59. To Draw a Square about a Given Circle.— In 
Fig. 200, let be the center of the given circle. Place 




[M D 



= / 



0/ 



c 



Fig. SOI.— To Draw a Square upon a Given Side. 

the blade of the T-square against the point 0, and 
draw the line A B. With one of the shorter sides 
E F, of a 45-degree set-square against the T-square, 
and with the other short side against the point 0, draw 
the line DOC. Move the T-square upward until it 
strikes the point C, and draw the line H C I. Move 
it down until it strikes the point D, and draw the line 
E D K. With the side E F of the set-square against 
the T-square, as shown in the engraving, bring the side 
E G against the point A, and draw E A H. In like 
manner bring it against the point B, and draw K B I, 
thus completing the figure. It is to be observed that 
the several lines composing the sides of the square are 
tangent to the circle in the points A C B and D re- 
spectively. The only object served by drawing the 
diameters A B and C D is that of obtaining greater 
accuracy in locating the points of tangency. 

60. To Draw a Square upon a Given Side.— Let A B 
of Fig. 201 be the given side placed parallel to one 



side of the drawing board. Place one of the shorter 
edges of a 45-degree set-square against the J-square, as 
placed for drawing the given side, and slide it along 
until the long edge touches the point A, and draw the 
diagonal line A C indefinitely. Place the T-square so 
that its head comes against the left side of the board, 
as shown by the dotted lines in the engraving, and, 
bringing the blade against the point A, draw A D in- 
definitely. Then bringing the blade against the point 
B, draw B C, stopping this line at the point of inter- 
section with the line A C, as shown at C. Bring the 
T-square back to the original position and draw the 
line C D, thus completing the figure. In the case of 
a large drawing board, unless the figure is to be located 
very near one corner of it, or in the case of a drawing 
board of which the adjacent sides are not at right 
angles, it will be desirable to use the right angle of the 
set-square, instead of changing the T-square from one 
side to the other, as above described. The object of 
drawing the diagonal line A C is to determine the 
length of the side C B. This also may be done by the 
use of the compasses instead of the set-square, as fol- 
lows : From B as center, with B A as radius, describe 
the arc A O C. Place the T-square as shown by the 
dotted lines, and, bringing it against the point B, draw 
B C, producing it until it intercepts the arc A O C in 
the point C. The remaining steps are then to be taken 
in the manner above described. 

III.— BY MEANS OF THE PROTRACTOR. 

The protractor, which has been already described 
and illustrated (see Fig. 116, Chapter II), is an instru- 
ment for measuring angles. The usual form of this 
instrument is a semicircle with a graduated edge, the 
divisions being more or less numerous, according to its 
size. In instruments of ordinary size the divisions are 
single degrees, numbered by 5s or by 10s, while in 
larger sizes the divisions are made to fractions of 
degrees. 

Since the protractor by its construction affords the 
means of measuring or of setting off any angle whatso- 
ever, it is especially useful in circumscribing or in- 
scribing polygons, or of erecting them upon a given 
side. As its use is of infrequent occurrence among 
pattern draftsmen, only a few problems in inscribing 
will be given, which will be sufficient to enable the 
reader to apply it in other cases that may arise. 

61. To Inscribe an Equilateral Triangle within a Given 
Circle.— In Fig. 202, let O be the center of the given 
circle. Through O draw a diameter, as shown by 



58 



TJie New Metal Worker Pattern Book. 



COD. Place the protractor so that its center point 
shall coincide with 0, and turn it until the point mark- 
ing 60 degrees falls upon the line COD. Then mark 
points in the circumference of the circle corresponding 
to (zero) and 120 degrees of the protractor, as shown 
by B and E respectively. Draw the lines C E, E B 
and B C, thus completing the required figure. The 
reasons for these several steps are quite evident. The 
circle consists of 360 degrees. Then each side of an 
equilateral triangle must represent one-third of 360 
degrees, or 120 degrees. Assume the point C for one 
of the angles, and draw the line COD. Then, by the 
nature of the figure to be drawn, D must fall opposite 
the center of one side. Therefore, since 60 is the half 
of 120 (the length of one side in degrees), place 60 
opposite the point D, and mark and 120 for the other 
angles, then complete the figure by drawing the lines 




Fig. SOS. — To Inscribe an Equilateral Triangle within a 
Given Circle. 

as shown. Since in many cases the protractor is much 
smaller than the circle in which the figure is to be con- 
structed, it becomes necessary to mark the points at 
the edge of the instrument, and carry them to the cir- 
cumference by drawing lines from the center of the 
circle through the points, producing them until the 
circle is reached. 

62. To Inscribe a Square within a Given Circle.— In 
Fig. 203, let be the center of the given circle. 
Through draw a diameter, as shown by C D. 
Place the protractor so that its center point coincides 
with 0, and turn it until the point marking 45 degrees 
falls upon the line COD. Mark points in the circum- 
ference of the circle corresponding to 0, 90 and 180 
degrees of the protractor, as shown by F, G and E re- 
spectively. From G, through the center 0, draw G O 
H, cutting the circumference of the circle in the point 



H. Then E, G, F and H are the angles of the required 
figure, which is to be completed by drawing the sides 
E G, G F, F H and H E. Since the circle is composed 
of 360 degrees, one side of an inscribed square must 
represent one-fourth part of 360 degrees, or 90 degrees. 
The half of 90 degrees is 45 degrees. Hence, in set- 
ting the protractor, the point representing 45 degrees 
was placed opposite the point in which it is desired the 
center of one of the sides shall fall, or, in other words, 
upon the line COD. Then, having marked points 90 
degrees removed from each other, or, as explained 
above, opposite the points 0, 90 and 180 of the pro- 
tractor, as shown by F, G and E, the fourth point was 
obtained by the diagonal line. It is evident that H 
must fall opposite G, upon a line drawn through the 
center. Or the protractor might have been moved 
around, and a space of 90 degrees measured from either 




Fig. SOS.— To Inscribe a Square within a Given Circle. 

F or E, which, as will be clearly seen, would have 
given the same point, H. 

63. To Inscribe an Octagon within a Given Circle.— 

Through the center O of the given circle, Fig. 204, 
draw a diameter, A O B, upon which the center of 
one side is required to fall. Place the protractor so 
that its center point shall coincide with the center O, 
and turn it so that the point representing 22£ degrees 
shall fall on the line A O B. Then mark points in 
the circumference of the circle corresponding to 0, 45, 
90, 135 and 180 degrees of the protractor, as shown 
by E, G, H, I and F. Beverse the protractor, and in 
like manner mark the points M, L and K ; or these 
points may be obtained by drawing lines from I, H 
and G respectively through the center O, cutting the 
circumference in M, L and K. The figure is to be 
completed by drawing the sides F I, I H, H G, G E, 



Geometrical Problems. 



59 



E M, M L, L K and K F. Since the circle consists 
of 360 degrees, one side of an octagon must represent 
45 degrees, or one-eighth of 360. The half of 45 is 
22-J-. Hence, the point of the protractor representing 
22-J- degrees was placed upon the line A B, which 
represents the center of one side of the required figure. 
Having thus established the position of one side, the 
other sides of the figure are located by marking points 
in the circumference of the circle opposite points in the 
protractor at regular intervals of 45 degrees. 

64. To Inscribe a Dodecagon within a Given Circle.— 
In Fig. 205, let be the center of the given circle. 




Fig. W4.—T0 Inscribe an Octagon within a Given Circle. 

Through draw the diameter A B, at right angles 
to which one of the sides of the polygon is required to 
be. Set the protractor so that the center point of it 
coincides with the center 0, and revolve it until the 
point marking 15 degrees falls upon the line A B. 
With the protractor in this position, mark points in 
the circumference of the circle opposite the points in 
the protractor representing 0, 30, 60, 90, 120, 150 and 
180 degrees, as shown by E, F, G, H, I, K and L. 
Then these points will represent angles of the required 
polygon. The remaining angles may be obtained by 
placing the protractor in like position in the opposite 
half of the semicircle, or they may be determined by 
drawing lines from the points F, Gr, H, I and K 
through the center O, producing them until they cut 



the circumference in the points M, N, P, R and S, 
which are the remaining angles. The figure is now to 
be completed by drawing the sides, as shown. In a do- 
decagon, or twelve-sided figure, each side must occupy 
a space represented by one-twelfth of 360 degrees, or 
30 degrees of the protractor. As the side F E was re- 
quired to be located in equal parts upon opposite sides 
of A O B, the middle of one division of the protractor 
representing a side (that is, 15 degrees, or one-half of 
30 degrees) was placed upon the line A O B. Having 
thus established the position of one side, the others 
are measured off in manner above described. 




p R 

Fig. S05. — To Inscribe a Dodecagon within a Given Circle. 

In making use of the protractor to erect a regular 
polygon upon a given side, the exterior angle, or angle 
formed by an adjacent side with the given side ex- 
tended, as E B D in Figs. 168, 170 and 171, is found 
by dividing 360 degrees by the number of sides 
in the required polygon ; while the interior angle, or 
angle between any two adjacent sides on the inside of 
the polygon, as E B A in the same diagrams, is the 
supplement of that angle, or, in other words, is found 
by subtracting the exterior angle from 180 degrees. 
Thus to find the exterior angle by means of which to 
construct a regular decagon, divide 360 degrees by 10; 
which gives 36 degrees; while the interior an<de is 
equal to 180 degrees less 36 degrees, which is 144 de- 
grees. 



THE ELLIPSE. 



For a definition of the ellipse the reader is referred 
to Chapter I, definitions 78 and 113. It may also be 
described as a curve drawn with a constantly increas- 
ing or diminishing radius, or as similar to a circle, but 
having one diameter longer than another, the diameters 
referred to being at right angles to each other. 



If, upon one of two lines intersecting each other 
at right angles, half of the long diameter be set off each 
way from their intersection (A A, Fig. 206) and upon 
the other line half of the short diameter be set off each 
way from the intersection (B B, Fig. 206), four prin- 
cipal points in the circumference of the ellipse will thus 



60 



Tlie New Metal Worker Pattern Book. 



be established ; and through these four points only- 
one perfect ellipse can be drawn, one-quarter of which 
is shown by the solid line from A to B in the illustra- 
tion. It is true that other curves having the appear- 
ance of an ellipse can be drawn through these points, 
as shown by the clotted lines, but, as stated above, 
there is only one curved line existing between those 
points which can be correctly termed an ellipse. 

There are several methods of producing a correct 
ellipse, as by a string and pencil, by a trammel con- 
structed for the purpose and by projection from an 
oblique section of a cylinder or of a cone, each of 
which will be considered in turn. The ellipse is 
properly generated from two points upon its major 
axis, called the foci, and its circumference is so drawn 
that if from any point therein two lines be drawn to 
the two foci, their sum shall be equal to the sum of 




Fig. 206. — Defining an Ellipse. 

two lines drawn from any other point in the circum- 
ference to the foci. 

65. To Draw an Ellipse to Specified Dimensions 
with a String and Pencil.— In Fig. 207, let it be re- 
quired to draw an ellipse, the length of which shall be 
equal to the line A B, and the width of which shall 
be equal to the line D C. Lay off A B and DC at 
right angles to each other, intersecting at their middle 
points, as shown at E. Set the compasses to one- 
half the length of the required figure, as A E, and 
liom either D or C as center, strike an arc, cuttino- 
A B in the points F and G. These points, F and G, 
then are the two foci, into which drive pins, as shown. 
Drive a third pin at C. Then pass the string around 
the three points F, Gr and C and tie it. Remove the 
pin C and substituting for it a pencil, pass the same 
around, as shown at P, keeping the string taut. If 
the combined lengths from F and Gr to the several 
points in the boundary line be set off upon a straight 



line, their sums will be found equal. For example, 
the sums of P F and P G, A F and A G, C F and 
C G, B F and B G, are all the same. 

Although correct so far as theory is concerned, 
this method is liable to error on account of the stretch- 
ing of the string. The same result can be obtained 
by means of a trammel constructed for the j^urpose, 
which is shown in Fig. 208. E is a section through 
the arms, showing the groove in which the heads of the 
bolts move. H and G are the bolts or pins by which 
the movement is controlled and regulated. In the 
engraving the bar K is shown with holes at fixed dis- 
tances, through which the governing pins are passed. 
An improvement upon this plan of construction con- 
sists of a device that will clamp the pins firmly to the 
bar at any point, thus providing for an adjustment of 
the most minute variations. 




Fig. 207.— To Draw an Ellipse by Means of a String and Pencil. 

66. To Draw an Ellipse to Given Dimensions by Means 
of a Trammel. — In Fig. 208, let it be required to de- 
scribe an ellipse, the length of which shall be equal to 
A B and the breadth of which shall be C D. Draw A 
B and C D at right angles, intersecting at their mid- 
dle points. Place the trammel, as shown in the en- 
o-ravine, so that the center of the arms shall come 
directly over the lines. First place the rod along the 
line A B, so that the pencil or point I shall coincide 
with either A or B. Then place the pin G directly 
over the intersection of A B and C D. Next place 
the rod along the line C D, bringing the pencil or 
point I to either C or D, and put the pin H over the 
intersection of A B and C D. The instrument is then 
ready for use, and the curve is described by the pencil 
I moved by the hand, and controlled by the pins work- 
ing in the grooves. 

"When a trammel is not convenient, a very fair 
substitute is afforded by the use of a common steel 



Geometrical Problems, 



61 



square and a thin strip of wood, like a lath. This 
method of drawing an ellipse is useful under ordinary 
circumstances when only a part of the figure is re- 
quired, as in the shape of the top of a window frame 
to which a cap is to be fitted, in which half of the 
figure would be employed, or in the shaping of a 
member of a molding in which a quarter, or less than 
a quarter, of the figure would be used. 

67. To Draw an Ellipse of Given Dimensions by 
Means of a Square and a Strip of Wood.— In Fig. 209, 
set off the length of the figure, and at right angles to 
it, through its middle point, draw a line representing 
the width of the figure. Place a square, as shown by 
A E C, its inner edge corresponding to the lines. Lay 
the strip of wood as shown by F E, putting a pencil at 
the point F, corresponding to one end of the figure, 



placed upon a board, and a line drawn around it, the 
resulting figure will be a circle. If now the pipe be cut 
obliquely, as in making an elbow at any angle, and the 
end thus cut be placed upon a board and a line drawn 
around it, as mentioned in the first case, the figure 
drawn will be an ellipse. What has thus been roughly 
done by mechanical means may be also accomplished 
upon the drawing board in a very simple and ex- 
peditious manner. The demonstration which follows 
is of especial interest to the pattern cutter, because 
the principles involved in it lie at the root of many 
practical operations which he is called upon to perform. 
For example, the shape to cut a piece to stop up the 
end of a pipe or tube which is not cut square across, 
the shape to cut a flange to fit a pipe passing through 
the slope of a roof, and other similar requirements of 




Fig, 208. — To Draw an Ellipse by 
Means of a Trammel. 




Fig. aw. Kg . m 

To Draw an Ellipse by Means of a Square and a Strip of Wood. 



and a pin at E, corresponding to the inner angle of the 
square. Then place the stick across the figure, as 
shown in Fig. 210, making the pencil, F, correspond 
with one side of the figure, and put a pin at Gr, corre- 
sponding with the inner angle of the square. Now 
move the stick from one position to the other, letting 
the points E and G slide, one against the tongue and 
the other against the blade of the square. The pencil 
point will then describe the required curve. In draw- 
ing the figure the square must be changed in position 
for each quarter of the curve. As shown in the en- 
gravings, it is correct for the quarter of the curve rep- 
resented by F D, Fig. 209. It must be changed for 
each of the other sections, its inner edge being brought 
against the lines each time, as shown. 

One definition of an ellipse is " a figure bounded 
by a regular curve, which corresponds to an obbque 
section of a cylinder." 

This can be practically illustrated by assuming a 
piece of stove pipe as the representative of the cylin- 
der. If the piece of pipe is cut square across, the end 



almost daily occurrence, depend entirely upon the 
principles here explained. 

68. To Describe the Form or Shape of an Oblique 
Section of a Cylinder, or to Draw an Ellipse as the 
Oblique Projection of a Circle.— The two propositions 
which are stated above are virtually one and the same 
so far as concers the pattern cutter, and they may be 
made quite the same so far as a demonstration is con- 
cerned. The explanation of the engraving is confined 
to the idea of the cylinder, believing it in that shape 
to be of more practical service to the readers of this 
book than in any other. In Fig. 211, let GEFH 
represent any cylinder, and A B C D the plan of the 
same. Let I K represent the plane of any oblique cut 
to be made through the cylinder. It is required to 
draw the shape of the section as it would appear if 
the cylinder were cut in two by the plane I K, and 
either piece placed with the end I K flat upon paper 
and a line scribed around it. Divide one-half of 
the plan ABC into any convenient number of equal 
parts, as shown by the figures 1,2,3,4, etc. Through 



62 



The New Metal Worker Pattern Booh. 



these points and at right angles to the diameter A 
C draw lines as shown, cutting the opposite side of 
the circle. Also continue these lines upward until 
they cut the oblique line I K, as shown by I 1 , 2 1 , 3 1 , 
etc. Draw I' K 1 , making it parallel to I K for con- 
venience in transferring spaces. With the 7- sc F iare 
set at right angles to I K, and brought successively 
against the points in it, draw lines through I 1 K 1 , 
as shown by l 2 , 2 2 , 3 2 , etc. With the dividers take 
the distance across the plan A B C D on each of 




Fig. 211. — The Ellipse as an Oblique Section of a Cylinder. 

the several lines drawn through it, and set the same 
distance off on corresponding lines drawn through 
F K 1 . In other words, taking A C as the base for 
measurement in the one case and F K 1 as the base of 
measurement in the other, set off from the latter, on 
each side, the same length as the several lines measure 
on each side of A C. Make 2 2 equal to 2, and 3 s 
equal to 3, and so on. Through the points thus ob- 
tained trace a line, as shown by I' M K' and the 
opposite side, thus completing the figure. 

To make this problem of practical use it is neces- 
sary that the diameter of the cylinder shall be equal 
to the short diameter of the required ellipse, and that 



the line I K be drawn at such an angle that the dis- 
tance l 1 9 1 shall be equal to its long diameter. 

Another definition of the ellipse is that "it is a 
figure bounded by a regular curve, corresponding to 
an oblique section of a cone through its opposite sides." 
It is this definition of the ellipse that classes it among 
what are known as conic sections. It is generally a 
matter of surprise to students to find that an oblique 
section of a cylinder, and an oblique section of a cone 
through its opposite sides, produce the same figure, 
but such is the case. The method of drawing an 
ellipse upon this definition of it is given in the follow- 




Fig. 212.— The Ellipse as an Oblique Section of a Cone. 

ing demonstration. The principles upon which this 
rule is based, no less than those referred to in the last 
demonstration, are of especial interest to the pattern 
cutter, because so many of the shapes with which he 
has to deal owe their origin to the cone. 

69. To Describe the Shape of an Oblique Section of a 
Cone through its Opposite Sides, or to Draw an Ellipse as 
a Section of a Cone.— In Fig. 212, let B A C represent 
a cone, of which E D G F is the plan at the base. 
Let H I represent any oblique cut through its opposite 
sides. Then it is required to draw the shape of the 
section represented by H I, which will be an ellipse. 
At any convenient place outside of the figure draw a 
duplicate of H I parallel to it, upon which to construct 
the figure sought, as H 1 I 1 . Divide one-half of the 



Geometrical Probteitts. 



63 



plan, as E D G, into any convenient number of equal 
parts, as shown by 1, 2, 3, 4, etc. From the center 
of the plan M draw radial lines to these points. From 
each of the points also erect a perpendicular line, which 
produce until it cuts the base line B C of the cone. 
From the base line of the cone continue each of these 
lines toward the apex A, cutting the oblique line H I. 
Through the points thus obtained in H I, and at right 
angles to the axis A D of the cone, draw lines, as 
shown by l 1 , 2', 3 1 , 4 1 , etc., cutting the opposite sides 
of the cone. From the same points in H I, at right 
angles to it, draw lines cutting H 1 I', as shown by 
l 2 , 2 2 , 3", 4 2 , etc., thus transferring to it the same 
divisions as have been given to other parts of the fig- 
ure. After having obtained these several sets of lines, 
the first step is to obtain a plan view of the oblique 
cut, for which proceed as follows : With the di- 
viders take the distance from the axial line AD to 
one side of the cone, on each of the lines l 1 , 2 1 , 3 1 , 4', 
etc., and set off like distance from the center of the 
plan M on the corresponding radial lines 1, 2, 3, 4, 
etc. A line traced through the points thus obtained 
will give the plan view of the oblique cut, as shown by 
the inner line in the plan. 

This result may be verified by dropping lines 
vertically from the points in H I across the plan, in- 
tersecting them with the radial lines in the plan of 
corresponding number. Thus a line dropped from 
point 4 on H I should intersect the radial line M 4 at 




Fig. SIS, 



-To Construct an Ellipse from Two Circles by 
Intersecting Lines. 



the same point (4 s ) established upon it by measuring 
the distance upon line 4 l from A D to A B. Having 
thus obtained the shape of the oblique cut as it would 
appear in plan, the next step is to set off upon the 
lines previously drawn through H 1 I 1 the width of the 
oblique cut in plan as measured upon lines of corre- 
sponding number. Therefore, with E G as a basis of 



measurement, with the dividers take the distance on 
each of the several cross lines 2 3 , 3 3 , 4 3 , 5 3 , etc., from 
E G to one side of the plan of the oblique cut just de- 
scribed, and set off the same distance on each side of 
II' I 1 on the corresponding lines. A line traced 
through the points thus obtained will be an ellipse. 




Fig. 211,.— To Draw an Ellipse within a Given Rectangle by 
Means of Intersecting Lines. 

70. To Construct an Ellipse to Given Dimensions by 
the Use of Two Circles and Intersecting; Lines.— In Fig. 
213, let it be required to construct an ellipse, the 
length of which shall equal A B and the width of 
which shall equal II F. Draw A B and H F at right 
angles, intersecting at their middle points, K. From 
K as center, and with one-half of the length A B as 
radius, describe the circle A C B D. From K as 
center, and with one-half of the width H F as radius, 
describe the circle E F G H. Divide the larger circle 
into any convenient number of equal parts, as shown 
by the small figures 1, 2, 3, 4, etc. Divide the smaller 
circle into the same number of equal and correspond- 
ing parts, as also shown by figures. By means of the 
T-square, from the points in the outer circle draw 
vertical lines, and from points in the inner circle draw 
horizontal lines, as shown, producing them until they 
intersect the lines first drawn. A line traced through 
these points of intersection will be an ellipse. 

71. To Draw an Ellipse within a Given Rectangle by 
Means of Intersecting- Lines.— In Fig. 214, let E D B 
A be any rectangle within which it is required to con- 
struct an ellipse. Bisect the end A E, obtaining the 
point F, from which erect the perpendicular F G, 
dividing the rectangle horizontally into two equal por- 
tions. Bisect the side A B, obtaining the point DZ, 
and draw the perpendicular H I, dividing the rectangle 
vertically into two equal portions. The lines F G and 
H I are then the axes of the ellipse. F G represents 
the major axis, and H I the minor axis. Divide the 
spaces F E, F A, G D and G B into any convenient 
number of equal parts, as shown by the figures 1, 2, 3. 
From these points in F E and G D draw lines to I, and 
from the points in F A and G B draw lines to the 



64 



Tlie Neio Metal Worker Pattern Book. 



point H. Divide F C and G C also into the same 
number of equal parts, as shown by the figures, and 
from H and I through each of these points draw lines, 
continuing them till they intersect lines of correspond- 
ing number in the other set, as indicated. A line 
traced through the several points of intersection be- 
tween the two sets of lines, as shown in the engraving, 
will be an ellipse. 

Besides the above methods for drawing correct 
ellipses there are several methods for drawing figures 
approximating ellipses more or less closely, but corn- 




Fig. 215- First Method. 




Fig. 216.— Second Method. 

To Draw an Approximate Ellipse with the Compasses, the 
Length only Being Given. 



posed of arcs of circles, which it is sometimes necessary 
to substitute for true ellipses for constructive reasons. 
The ellipse has been described above as a curve drawn 
with a constantly changing radius. If, instead of 
using an infinite number of radii, some finite number 
be assumed, it will appear that the greater the number 
assumed the more nearly will it approach a perfect 
ellipse. Thus, a curve very much like an ellipse can 
be drawn, each quarter of which is composed of arcs 
drawn from two centers. If the number of centers be 
increased to three, the curve comes much nearer a true 



ellipse, and with four or five centers to each quarter, 
the curve thus produced can scarcely be distinguished 
from the perfect ellipse. 

12. To Draw an Approximate Ellipse with the Com- 
passes, the Length only being Given.— In Fig. 215, let 
A C be any length to which it is desired to draw an 
elliptical figure. Divide A C into four equal parts. 
From 3 as center, and with 3 1 as radius, strike the arc 
BID, and from 1 as center, and the same radius, 
strike the arc B 3 D, intersecting the arc first struck 
in the points B and D. From B, through the points 1 
and 3, draw the lines B E and B F indefinitely, and 
from D, in like manner, draw the lines D G and D H. 
From the point 1 as center, and with 1 A as radius, 
strike the arc E G, and from 3 as center, with the same 
radius, or its equivalent, 3 C, strike the arc H F. 
From D as center, with radius D G, strike the arc G H, 
and from B as center, with the same radius, or its 
equivalent, B E, strike the arc E F, thus completing 
the figure. 

A figure of different proportions may be drawn in 
the same general manner as follows : Divide the length 
A C into four equal parts, as indicated in Fig. 216. 
From 2 as center, and with 2 1 as radius, strike the 
circle 1 E 3 F. Bisect the given length A C by the 
line B D, as shown, cutting the circle in the points E 
and F. From E, through the points 1 and 3, draw the 
lines E G and E H indefinitely, and from F, through 
the same points, draw similar lines, F I and F K. 
From 1 as center, and with 1 A as radius, strike the 
arc I A G, and from 3 as center, with equal radius, 
strike the arc K C H. From E as center, and with 
radius E G, strike the arc G D H, and from F as center, 
with corresponding radius, strike the arc I B K, thus 
completing the figure. 

73. To Draw an Approximate Ellipse with the Com- 
passes to Given Dimensions, Using Two Sets of Centers.— 
First Method. — In Fig. 217, let A B represent the 
length of the required figure and D E its width. 
Draw A B and D E at right angles to each other, and 
intersecting at their middle points. At the point A 
erect the perpendicular A F, and in length make it 
equal to C D. Bisect A F, obtaining the point N. 
Draw N D. From F draw a line to E, as shown, cut- 
tine N D in the point G. Bisect the line G D by the 
line H I, perpendicular to G D and meeting D E in the 
point I. In the same manner draw lines correspond- 
ing to G I, as shown by L I, M and B 0. From I 
and O as centers, and with I G as radius, strike the 
arcs G D L and MEE, and from K and P as centers, 



Geometrical Problems. 



65 



with K G as radius, strike the arcs GAM and L B E, 
thus completing the figure. 

74. To Draw an Approximate Ellipse with the Com- 
passes to Given Dimensions, Using two Sets of Centers. 
— Second Method. — In Fig. 218, let C D represent the 
length of a required ellipse and A B the width. Lay 
off these two dimensions at right angles to each other, 
as shown. On C D lay off a space equal to the width 
of the required figure, as shown by D E. Divide the 
remainder of D C, or the space E C, into three equal 
parts, as shown in the cut. With a radius equal to 
two of these parts, and from B as center, strike the 
circle G S F T. Then with F and G as centers, and F G 
as radius, strike the arcs, as shown, intersecting upon 
A B prolonged at and P. From 0, through the 
points G and F, draw L and M, and likewise from 



center, with F L as radius, describe a circle, as shown, 
thus establishing the points M, N and 0, which, with 
L, are the centers from which the ellipse is to be struck. 
From M, draw M L Q and M N S indefinitely, and in 
a similar manner L P and ONE. With as center, 
and D as radius, strike the arc P D E, cutting P 
and E, as shown. In a similar manner, and with 
the same radius (or which is the same, with M E as 
radius) and M as center, describe the arc Q E S. 
With L and N as centers, and with L B or N C as 
radius, strike the arcs Q B P and ECS, thus com- 
pleting the figure. 

The above methods of drawing approximate el- 
lipses are only available within certain limits of pro- 
portion, as will be discovered if an attempt is made 
to draw them very much elongated, the limit being 






e *• E 

Fig. 217.— First Method. Fig. 218.— Second Method. Fig. 219.— Third Method. 

To Draw an Approximate Ellipse with the Compasses, Using Two Sets of Centers. 



P, through the same points, draw P K and P N. 
From as center, with A as radius, strike the arc 
L M, and with the same radius, and P as center, strike 
the arc K N. From F and G as centers, and with F 
D and G as radii, strike the arcs N M and K L re- 
spectively, thus completing the figure. 

15. To Draw an Approximate Ellipse with the Com- 
passes to Given Dimensions, Using: Two Sets of Centers.— 
Third Method. — In Fig. 219, let B C represent the length 
of the required figure and D E its width. B C and D E 
are drawn at right angles to each other, intersecting at 
their middle points at F. The next step in describing 
the figure is to obtain the difference in length between 
the axes F D and F B, which can be done as indicated 
by the are D G. This difference, G B, is to be set off 
from the center F on F B and F D, as shown by F H, 
F J, then draw H J and set off half of H J to L, as 
indicated by the arc K L. The object of the operation 
so far has been to secure the point L. From F as 



reached when the long diameter is about equal to two 
times the shorter diameter. Beyond this limit in the 
first two methods, if the final arc be drawn with the 
radius G K (Figs. 217 and 218), it will not reach the 
end of the long diameter, but will strike it at a point 
inside of A or C. By the third method, if the long 
diameter be increased until it is about 2| times the 
shorter, the point L (Fig. 219) will fall at the extreme 
limit of the long diameter (B), thus completely cutting 
out the small arc P Q. It must, therefore, in extreme 
cases be left to the judgment of the draftsman to 
adjust or vary the lengths of the radii of the two arcs, 
so as to produce the result which will look the best. 

76. To Draw an Approximate Ellipse with the Com- 
passes to Given Dimensions, Using: Three Sets of Centers. 
—In Fig. 220, let A B represent the length of the re- 
quired figure and D E the width. Draw A B and D 
E at right angles to each other, intersecting at their 
middle points, as shown at C. From the point A draw 



66 



The New Metal Worker Pattern Book. 



A F, perpendicular to A B, and in length equal to C 
D. Join the points F and D, as shown. Divide A F 
into three equal parts, thus obtaining the points Z and 
I, and draw the lines Z D and I D. Divide A C into 
three equal parts, as shown by Y and G, and draw E 
G and E Y, prolonging them until they intersect with 
Z D and I D respectively, in the points J and H. 
Bisect J D, and draw K L perpendicular to its central 
point, intersecting D E prolonged in the point L. Draw 
J L and H J. Bisect H J, and draw M N perpen- 
dicular to its central point, meeting J L in N. Draw 
N H, cutting A B in the point 0. L then is the 
center of the arc J D P, N is the center of the arc H 
J, and is the center of the arc H A B. The points 
S and U, corresponding to N and 0, from which to 



of N 0, draw P R, perpendicular to N O and parallel 
to K M. Then N" and P R are the axes of the 
ellipse. 

78. In a Given Ellipse, to Find Centers by which an 

Approximate Figure may be Constructed In Fig. 222, 

let A E B D be any ellipse, in which it is required to 
find centers by which an approximate figure may be 
drawn with the compasses. Draw the axes A B and 
E D. From the point A draw A F, perpendicular to 
A B, and make it equal to C E. Join F and E. 
Divide A F into as many equal parts as it is desired 
to have sets of centers for the figure. In this in- 
stance four. Therefore, A F is divided into four 
equal parts, as shown by P and G. Divide A C 
into the same number of equal parts, as shown by R 




Fig. 320.— To Draw an Approximate 
Ellipse with the Compasses, Usiny Three 
Sets of Centers. 



K 












yy 












// 












»/'' 










\e 


1 h\ 
t i 


- — 


/f 






A 

/ \ 


1 / 






---._ 


c 


/ 


\ Ll / 






~~--— . 


~-~- / / 


\v 










/ -/ 
1 7 I 


AX \ 










/ / 
/ / 










/^-^ 




M 











Fig. 221. — To Find the True Axes of a 
Given Ellipse. 




Fig. 222.— In a Given Ellipse, to Find 
Centers by which an Approximate Figure 
may be Constructed. 



strike the remainder of the upper part of the figure, 
may be obtained by measurement, as indicated. Hav- 
ing drawn so much of the figure as can be struck from 
these centers, set the dividers to the distance L P or L 
J, and placing one point at E, the remaining center 
will be found at the other point of the dividers, in the 
line E D prolonged, as shown by X. 

XI. To Find the True Axes of a Given Ellipse.— In 
Fig. 221, let N P R be any ellipse, of which it is re- 
quired to find the two axes. Through the ellipse draw 
any lines, A B and D E, parallel to each other. Bisect 
these two lines and draw F G, prolonging it until it 
meets the sides of the ellipse in the points H and I. 
Bisect the line H I, obtaining the point C. From C as 
•center, with any convenient radius, describe the arc 
K L M, cutting the sides of the ellipse at the points K 
and M. Join K and M by a straight line, as- shown. 
Bisect M K by the line N 0, perpendicular to it. 
Through C, which will also be found to be the center 



S T. From the points of division in A F draw lines 
to E. From D draw lines passing through the divi- 
sions in A C, prolonging them until they intersect the 
lines drawn from A F to E, as shown by D U, D V 
and D W. Draw the chords U V, V W and W E, 
and from the center of each erect a perpendicular, 
which prolong until they intersect as follows : The 
line perpendicular to WE intersects the center line 
E D in the point D. Now draw D W and prolong 
the perpendicular to V W till it intersects D W in K, 
and draw K V. Prolong the perpendicular to U V 
till it cuts K V in L and draw L U, cutting A C in the 
point S. Then D is the center of the arc E "W, K is 
the center of the arc W V, L is the center of the arc 
V U and S is the center of the arc U N. By these 
centers it will be seen that one-quarter of the figure 
(A to E) may be struck. By measurement, corre- 
sponding points may be located in other portions of 
the figure. II correctly done the points U, V and W 



Geometrical Problems. 



67 



will be found to fall upon the ellipse, consequently 
the arcs drawn between those points from the centers 
obtained cannot deviate much from the correct ellipse. 
79. To Draw the Joint Lines of an Elliptical Arch.— 
First Method.— In a circular arch the lines representing 
the joints between the stones forming the arch, or the 
voussoirs as they are properly called, are drawn 
radially from the center of the semicircle of the arch. 
In an elliptical arch this operation is somewhat more 
difficult as the true ellipse possesses no such single 




B c 

Fig 223.— First Method. 
To Draw the Joint Lines of an Elliptical Arch. 

point, but, instead, two foci, as has been explained. 
Therefore, the following course must be pursued : 
From any point upon the ellipse at which it is desired 
to locate a joint, as A, Fig. 223, draw a line to each 
of the foci, as A B and A C. Bisect the angle B A C 
(Prob. 12 in this chapter), as shown at D, and extend 
the line D A outside the ellipse, which will be the joint 
line required. 

80. To Draw the Joint Lines of an Elliptical Arch.— 

Second Method In Fig. 224, A B is one-half the curve 

of the arch, A C its center line and C B its springing 



line. Draw A D parallel to C B, and D B parallel to 
A C, and draw the diagonals A B and C D. From 
each of the points 1, 2, 3, etc., representing the joints, 
drop lines vertically, cutting C D. From their inter- 
sections with C D carry them at right angles to A B, 
cutting the springing line C B, as shown by the small 
figures 1% 2% 3% etc. From the points in C B draw 













D 




I ' — ~- 
1' / 


3/ 


3 /// \ 


5*S 


1* 


2 2 i 


* i ! 5 


^ 


B 



Fig. 224.— Second Method. 
To Draw the Joint Lines of an Elliptical Arch. 

lines through corresponding points in the arch A B, 
as I s 1, 2 s 2, 3 s 3, etc., and continue them through 
the face of the arch which will be the joint lines 
sought. 

In the case of an elliptical curve made up of arcs 
of circles, the joint lines would be drawn radially from 
the centers of the arcs in which they occur. 



THE VOLUTE. 



The volute is an architectural figure of a geo- 
metrical nature based upon the spiral, and is of quite 
frequent occurrence in one form or another, conse- 
quently some remarks xipon the different methods of 
drawing it will not be out of place. 

81. To Draw a Simple Volute. — Let D A, in Fig. 
225, be the width of a scroll or other member for 
which it is desired to draw a volute termination. 
Draw the line D 1, in length equal to three times D 
A, as shown by D A, A B and B 1. From the point 
1 draw 1 2 at right angles to D 1, and in length equal 
to two-thirds the width of the scroll — that is, to two- 
thirds of D A. From 2 draw the line 2 3 perpen- 
dicular to 1 2, and in length equal to three-quarters of 
1 2. Draw the diagonal line 1 3. From 2 draw a 
line perpendicular to 1 3, as shown by 2 4, indefi- 
nitely. From 3 draw a line perpendicular to 2 3, pro- 



ducing it until it cuts the line 2 4 in the point 4. 
From 4 draw a line perpendicular to 3 4, producing 
it until it meets the line 1 3 in the point 5. In like 
manner draw 5 6 and 6 7. The points 1, 2, 3, 4, 
etc., thus obtained are the centers by which the curve 
of the volute is struck. From 1 as center, and with 1 
D as radius, describe the quarter circle D C. Then 
from 2 as center, and 2 as radius, describe the 
quarter circle C F, and so continue using the centers 
in their numerical order until the curve intersects with 
the other curve beginning at A and struck from the 
same centers, thus completing the figure, as shown. 

82. To Draw an Ionic Volute. — Draw the line A 
B, Fig. 226, equal to the hight of the required volute, 
and divide it into seven equal parts. From the third 
division draw the line 3 C, and from a point on this 
line at any convenient distance from A B describe a 



08 



The New Metal Worker Pattern Book. 



circle, the diameter of which shall equal one of the 
seven divisions of the line A B. This circle forms 
the eye of the volute. In order to show its dimen- 
sions, etc., it is enlarged in Fig. 227. A square, D 
E F G, is constructed, and the diagonals G E and F 
D are drawn. F E is bisected at the point 1, and the 
line 1 2 is drawn parallel to G E. The line 2 3 is 
then drawn indefinitely from 2 parallel to F D, cut- 
ting G E in the point H. The distance from H to the 
center of the circle is divided into three equal 
parts, as shown by H a b 0. The triangle 2 1 is 
formed. On the line II set off a point, as c, at a 
distance from O equal to one-half of one of the three 
equal parts into which H has been divided. From 
c draw the line c 3 parallel to 1 0, producing it until 
it cuts 2 3 in the point 3. From 3 draw the line 3 4 
parallel to G E indefinitely. From the point c draw a 
line c 4 parallel to 2 0, cutting the line 3 4 in the 
point 4, completing the triangle c 3 4. From 4 draw 
the line 4 5 parallel to F D, meeting 1 in the point 
5. From 5 draw the line 5 6 parallel to G E, meeting 
the line 2 in the point 6. From 6 draw the line 6 
7 parallel to F D, meeting the line c 3 in the point 7. 




Fig. 225.— To Draw a Simple Volute. 

Proceed in this manner, obtaining the remaining points, 
8, 9, 10, 11 and 12. These points form the centers 
by which the outer line of the volute proper is drawn. 
From 1 as center, and with radius 1 F, Fig. 226, de- 
scribe the quarter circle F G. Then from 2 as center, 
and with radius 2 G describe the quarter circle G D, 
and so continue striking a quarter circle from each of 
the centers above described until the last arc meets 



the circle first drawn. To obtain the centers by which 
the inner line of the volute is struck, and which 
gradually approaches the outer line throughout its 
course, proceed as follows : Produce the line 3 c, 
Fig. 227, until it intersects 1 2 in the point l l , which 




Fig. 226.— To Draw an, Ionic Volute. 

mark. This operation gives also the points 9 1 and 5' 
of intersection with the lines parallel to 1 2, which 
also mark. In like manner produce 4 c, 1 O and 2 
0, as shown by the dotted lines, and mark the several 
points of intersection formed with the cross lines. 
Then the points l 1 , 2 1 , 3', 4 1 , etc., thus obtained are 
the centers for the inner line of the volute, which 
use in the same manner as described for producing 
the outer line. 

83. To Draw a Spiral from Centers with Compasses. 
— Divide the circumference of the primary — some- 
times called the eye of the spiral — into any number 
of equal parts ; the larger the number of parts the 
more regular will be the spiral. Fig. 228 shows the 
primary divided into six equal parts. Fig. 229 is an 
enlarged view of this portion of the preceding figure. 
Construct the polygon by drawing the lines 1 2, 2 3, 
3 4, etc., producing them outside of the primary, as 
shown by A, B, D, F, C and E. From 2 as center, 
with 2 1 as radius, describe the arc A B. From 3 as 
center, and 3 B as radius, describe the arc B D ; and 



Geometrical Problems. 



69 



with, 4 as center, with radius 4 D, describe the arc ~D 
F. In this manner the spiral may be continued any 
number of revolutions. In the resulting figure the 
various revolutions will be parallel. 

84. To Draw a Spiral by Means of a Spool and 
Thread. — Set the spool as shown by ADB in Fig. 



/ 2 


















/'/\ \ 


/ A 2 


V\ 


' 


;'/ 










// 






6' 


\\ 


1 o 


¥/ 






.* 




/ A 




















\ H 


a 


b 










\ \ 




7/'" 


.8' 




\ \. 


/ 


/.' 




k\ 




\ \ 


A 


/7> 




4 / j 


\ >> 


/ / 






\ 












* / / 



Fig. 227— Eye of the Volute in Fig. 226 Enlarged. 

230 and wind a thread around it. Make a loop, E, 
in the end of the thread, in which place a pencil, as 
shown. Hold the spool firmly and move the pencil 
around it, unwinding the thread. A curve will be de- 
scribed, as shown in the dotted lines of the engrav- 
ing. It is evident that the proportions of the figure 



top, E A at the bottom and A B at the side, the 
length of A B, which determines the width of the 
scroll, being given. Bisect A B, obtaining the 
point C. Let the distance between the beginning 
and ending of the first revolution of the scroll, shown 
by a e, be established at pleasure. Having determined 




1 

4 , 




3 


8 




7 






5 


B 






1 


o 



Fig. 2Sl.—To Draw a Scroll to a 
Specified Width. 



Fig. 232.— The Center of Fig. 
231 Enlarged. 



this distance, take one-eighth of it and set it off up- 
ward from C on the line A B, thus obtaining the point 
b. From b draw a horizontal line of any convenient 
length, as shown by b h. With one point of the com- 
passes set at b, and with b A as radius, describe an arc 
cutting the line b h in the point 1. In like manner, 
from the same center, with radius b B, describe an arc 
cutting the line b h in the point 2. Upon 1 2 as a base 
erect a square, as shown by 1 2 3 4. Then from 1 as 





Fig. 228. — To Draw a Spiral from Centers. 



Fig. 229. — Enlarged View of the Eye of the Spiral in 
Fig. 228. 



Fig. 230.— To Draw a Spiral by 

Means of a Spool and Thread. 



wee determined by the size of the spool. Hence a 
larger or smaller spool is to be used, as circumstances 
require. 

85. To Draw a Scroll to a Specified Width, as for a 
Bracket or Modillion. — In Fig. 231, let it be required to 
construct a scroll which shall touch the line D B at the 



center, with 1 a as radius, describe an arc, a b ; and 
from 2 as center, with 2 b as radius, describe the arc 
b c. From 3 as center, with radius 3 c, describe the 
arc c d. From 4 as center, with radius 4 d, describe 
the arc d e. If the curve were continued from e, being: 
struck from the same centers, it would run parallel to 



70 



The New Metal Worker Pattern Book. 



itself ; but as the inner line of the scroll runs parallel 
to the outer line, its width may be set off at pleasure, 
as shown by a a', and the inner line may be drawn by 
the same centers as already used for the outer, and con- 
tinued until it is intersected by the outer curve. To 
find the centers from which to complete the outer 
curve, construct upon the line of the last radius above 
used (4 e) a smaller square within the larger one, as 
shown by 5 6 7 8. This is better illustrated by the 
larger diagram, Fig. 232, in which like figures repre- 
sent the same points. Make the distance from 5 to 8 



equal to one-half of the space from 4 to 1, making 
4 to 8 equal the distance of 5 to 1. Make 5 to 6 equal 
the distance from 8 to 5. After obtaining the points 
5, 6, 7, etc., in this manner, so many of them are to 
be used as are necessary to make the outer curve inter- 
sect the inner one, as shown at g. Thus 5 is used as 
a center for the arc ef, and 6 as a center for the arc 
f g. If the distance a a' were taken less than here 
given, it is easy to see that more of the centers upon 
the small square would require to be used to arrive at 
the intersection. 



"^- 



CHAPTER V. 



Pried pies of Pattern Ctatttaj 



To any one wishing to pursue pattern cutting as 
a profession it is essential not only that lie know how 
to solve a large number of intricate problems, but that 
lie understand thoroughly the principles which under- 
lie such operations. It is, therefore, appropriate, be- 
fore introducing pattern problems, that some attention 
should be given to the explanation of such principles 
in order that the reasons for the steps taken in the 
demonstrations following may be readily understood. 
Underlying the entire range of problems peculiar to 
sheet metal work are certain fundamental principles, 
which, when thoroughly understood, make plain and 
simple that which otherwise would appear arbitrary, 
if not actually mysterious. So true is this that noth- 
ing is risked in asserting that any one who thoroughly 
comprehends all the steps in connection with cutting 
a simple square miter is able to cut any miter what- 
soever. Since almost any one ean cut a square miter, 
the question at once arises, in view of this statement, 
why is it that he cannot cut a raking miter, or a pin- 
nacle miter, or any other equally difficult form? The 
answer is, because he does not understand how he cuts 
the square miter. He may perform the operation 
just as he has been taught, and produce results entirely 
satisfactory from a mechanical standpoint, without be- 
ing intelligent as to all that he has done. He does 
not comprehend the why and wherefore of the steps 
taken. Hence it is that when he undertakes some 
other miter he finds himself deficient. 

There is a wide difference between the skill that 
produces a pattern by rote — by a mere effort of the 
memory — and that which reasons out the successive 
steps. One is worth but very little, while the other 
renders its possessor independent. It is with a desire 
to put the student in possession of this latter kind of 
skill, to render him intelligent as to every operation 
to be performed, that the present chapter is written. 

The forms with which the pattern cutter has to 
deal may be divided, for convenience of description, 
into three general classes : 



I. The first of these embraces moldings, pipes 
and regular continuous forms, and may be called forms 
of parallel lines, or as a shorter and more convenient 
name to use, parallel forms. 

II. The second, which will be called regular taper- 
ing; forms, comprehends all shapes derived from cones 
or pyramids, or from solids having any of the regular 
geometric figures as a base and which terminate in an 
apex. 

HI. The third class will be called irregular forms, 
and will include everything not classified under either 
of the two previous heads. Many of these might be 
properly called transition pieces — that is, pieces which 
have figures of various outlines placed at various 
angles as their bases, and have figures with differing 
outlines variously placed, as their upper terminations, 
thus forming transitions, or connecting pieces between 
the form which lies next them at one end and the 
adjacent form on the other end. 

While pieces of metal of any shape necessary to 
form the covering of a solid of any shape may prop- 
erly be called patterns, the shapes of pieces necessary 
to form the joints between moldings meeting at an 
angle are known distinctively as miters. This name 
applies equally well in sheet metal work if the two 
arms of the molding are not of the same profile, or to 
a single arm coming against any plain or irregular sur- 
face. These forms comprise the first class referred to 
above and, so far as principle is concerned, come under 
the same general rules, which will be subsequently 
given. 

Conical forms, with very little taper, coming 
against other forms are also said to miter with them. 
In fact, the word miter has come into such general use 
that it is often applied to any joint between pieces of 
metal ; but the term can scarcely be considered as cor- 
rect when the forms have very much taper. The 
principle involved in the development of such patterns, 
however, is the same as that applied to the develop- 
ment of the surfaces of all other regular tapering forms, 



72 



The New Metal Worker Pattern Book. 



referred to above as the second class, whose character- 
istics will be considered in their proper chapter. 

The method employed for developing the patterns 
for forms of the third class has been termed triangula- 
tion, and is adopted on account of its simplicity, as it 
does away with the reduction or subdivision of an 
irregular form into a number of smaller regular forms, 
each one of which would have to be treated separately 
and perhaps by a different method. In fact, there are 
some shapes which have arisen from force of circum- 
stances which it would be impossible to separate into 
regular parts, and even if they could be so separated 
such a course would result in tedious and complicated 
operations. 

After principles have been thoroughly explained 
the problems in this work will follow in three sections 
or departments of the final chapter, arranged according 
to the above classification. 



This is one of the instances in which the pattern cut- 
ter is required to be something of an architectural 
draftsman, and to this end a chapter on Linear Draw- 
ing (Chap. Ill) has been introduced, in which atten- 
tion is given to this phase of the work, and to which 
the student is referred. 

The arrangement of the problems in each of the 
sections of the succeeding chapter will be made with 
reference to these two conditions, the simpler ones 
being placed before those in which preliminary draw- 
ing is required. 

Parallel FormtiSo 

(MITER CUTTING.) 

Since in sheet metal work a molding is made by 
bending the sheet until it fits a given stay, a molding 
may be defined mechanically as a succession of paral- 






Fig. 233.— Profile of a Molding. 



Fig. 234.— A Stay. 



Fig. 235.— A Reverse Stay. 



Two conditions exist in regard to the work of 
developing patterns of all forms, no matter to which 
of the three classes above defined they may belong : 

First — In very many cases a simple elevation or plan 
of the intersecting parts, together with their profiles,, is 
all that is necessary to begin the work of developing 
the pattern — that is, the plan or elevation, a3 the case 
may be, shows the line (either straight or curved) 
which represents the surface against which another 
part is to be fitted ; in other words, the much sought 
for ' ' miter line. ' ' 

Second — In numerous other instances, however, 
no view can be drawn either in elevation, oblique or 
otherwise, or in plan, in which the miter or junction 
1 of the parts will appear as a simple straight or curved 
line against which the points can be dropped. In such 
cases it becomes necessary to do some preliminary 
work in order to prepare the way to the actual work 
of laying out the pattern. A view of the joint must 
be developed by means of intersections of lines which 
will show it as it appears in connection with the eleva- 
tion or plan to be used in developing the pattern. 



lei forms or bends to a given stay, and, so far as the 
mechanic is concerned, any continuous form or ar- 
rangement of parallel continuous forms, made for any 
purpose whatever, may be considered a molding and 
treated under the same rules in all the operations of 
pattern cutting. Keejung this fact in mind all paral- 
lel forms will be considered as moldings and that 
word will be used in the demonstrations, remembering 
that a difference in name simply means a difference of 
profile, but not a difference in treatment or principle. 
A molding may be defined theoretically as a form 
or surface generated by a profile passed in a straight 
or curved line from one point to another, this profile 
being the shape that would be seen when looking at 
its end if the molding were cut off square. A prac- 
tical illustration of this may be given as follows : In 
Fig. 233, let the. form shown be the profile of some 
molding. If this shape be cut out of tin plate or 
sheet iron, as shown in Fig. 234, it is . called a stay. 
For the purpose of this illustration, as will appear fur- 
ther on, a stay, the reverse of the one shown in Fig. 
234, or, in other words, the piece cut from the face or 



Principles of Pattern Cutting. 



73 



outside of the shape represented in that figure, as 
shown in Fig. 235, will be required. 

Having made a reverse stay, or ' ' outside stay, ' ' 
as it is sometimes called, Fig. 235, take some plastic 
material — as potters' clay — and, placing it against any 
smooth surface, as of a board, place the stay against 
the board near one end in such a position that its ver- 
tical lines are parallel with the ends of the board, and 
move this reverse stay in a straight line along the face 
of the board until a continuous form is obtained in the 
clay corresponding to the profile of the stay, all as 
illustrated in Fig. 236. By this operation will be 
produced a molding in accordance with the second 
definition above given. The purpose in introducing 
this illustration is to show more clearly than is other- 
wise possible the principles upon which the different 




Fig. 236. — Generating a Molding in Plastic Material by Means of a Reverse Stay. 



parts of a molding are measured in the process of pat- 
tern cutting. 

Suppose that the form produced as illustrated in 
Fig. 236 be completed, and that both ends of the 
molding be cut off square. It is evident, upon in- 
spection, that the length of a piece of sheet metal 
necessary to form a covering to this molding will be 
the length of the molding itself, and that the width of 
the piece will be equal to the distance obtained by 
measuring around the outline of the stay which was 
used in giving shape to the molding. Now with a 
thin-bladed knife, or by means of a piece of fine wire 
stretched tight, let one end of the clay molding just 
constructed be cut off at any angle. By inspection of 
the form when thus cut, as clearly shown in the upper 
part of Fig. 237, it is evident that the end of a pattern 
to form a covering of this model must have such a 
shape as will make it when formed up conform to the 
oblique end of the molding or model. 



To cut such a pattern by means of a straight line 
drawn from a point corresponding to the end of the 
longer side of the mold, to a point corresponding to 
the end of the shorter side of it, would not be right, 
evidently, because certain parts of the covering, when 
formed up, fold down into the angles of the molding, 
and therefore would require to be either longer or 
shorter, as the case might be, than if cut as above de- 
scribed. It is plain, then, that some plan must be 
devised by which measurements can be taken in all 
these angles or bends, and at as many intermediate 
points as may be necessary, in order to obtain the right 
length at all points throughout its width. This can 
be done quite simply as follows : 

Divide the curved parts of the stay into any con- 
venient number of equal parts, and at each division 
cut a notch, or affix a point to it. 
Eeplace the stay in the position 
it occupied in producing the 
molding shown in Fig. 236 and 
pass it again over the entire 
length of the model. The points 
fastened to the stay will then 
leave tracks or lines upon the 
surface of the molding. Now, 
by means of measurement upon 
the different lines thus produced, 
the length of the molding at all of 
the several points established in 
the stay may be obtained. All 
this is clearly illustrated in Fig. 
237. In the upper right hand corner of the illus- 
tration is shown the stay prepared with points, by 
moving which as above described lines are left upon 
the face of the molding, as shown to the left 

Now, upon a sheet of paper fastened to a draw- 
ing board, draw a vertical line, as shown by A B in 
Fig. 237, and upon that line set off with the dividers 
the width of each space or part of the profile or stay — . 
that is, make the space 1 2 in the line A B equal to 
the space 1 2 in the stay, and 2 3 in the line A B 
equal to 2 3 of the stay, and so continue until all the 
spaces are transferred — and from the points thus ob- 
tained in A B draw lines at right angles to it indefi- 
nitely, as shown to the left. The lines an dspaces upon 
the paper will then correspond to the lines and spaces 
upon the clay molding made by the points fastened to 
the stay. Next, measure with the dividers the length 
of the molding upon each of the lines drawn upon it, 
and set off the same lengths upon the corresponding 



74 



The New Metal Worker Pattern Book. 



lines drawn upon the paper. This gives a series of 
points through which a line may be traced which will 
correspond in shape to the oblique end of the molding. 
Thus, set off from A B on the line 1 on the paper the 
length of the molding, measured from its straight end 
to its oblique end, upon the line produced by point 1 
'of the stay upon its face : and upon each of the other 
lines on the paper set off the length of the molding on 
the corresponding line on its face, meas- 
uring from the square end each time, 
which is represented by the line A B of 
the drawing. By this means are ob- 
tained points through which, if a line be 
traced, as shown by C D, the pattern of 
the covering will be described. The line 
A B, containing measurements from the 
profile, is called the " stretchout line," 
and the lines drawn through the points in 
it and at right angles to it are mathe- 
matically known as ordinates, but will in 
this work be called " measuring lines." 

Now, what has been clone in Fig. 
237 illustrates what is called." miter cut- 
ting," which in other words consists in 
describing upon a flat surface the shape of 
a given form or envelope, so that when 
the envelope is cut out of the flat sur- 
face and formed up to the stay from which 
its stretchout was derived, the finished 
molding will fit against a given surface at 
a given angle previously specified. 

The pattern shown in the lower part 
of Fig. 237, which has been obtained by 
means of a clay model, and measurements 
for which were obtained from the lines 
drawn on the surface of the clay model — 
maybe obtained just as well from a draw- 
ing. The question then is, how can the 
same results be obtained by lines drawn upon a flat 
surface as were obtained by measurements on lines 
drawn along the surface of a molding ? 

In moving the stay along the clay molding, cer- 
tain lines were made by means of the points affixed. 
If the reader will carefully examine Fig. 237 he will 
notice that the lines upon the molding made by this 
means corresponded in number and position with the 
points in the profile when it is laid flat on its side, in a 
position exactly opposite the end of the model, as shown. 

Hence, if the profile be drawn upon paper and in 
line with it, the . elevation terminated by the oblique 



line, which represents the surface against which it is 
required to miter, the same results can be accom- 
plished, care only being necessary that the relative 
positions of the parts be correctly maintained. 

This is illustrated in Fig. 238, which is to be 
compared with Fig. 237, and shows : First, that the 
profile A is drawn in correct position. Next, that 
from it the elevation F C D Gr of the molding is pro- 




Fig. 237.- 



The Use of Lines in Obtaining the Envelope of a Molding from a Model 
of the Same. 



jected, as follows : Use the T-square in the general 
position shown by B in the engraving, bringing it 
against the several points in A in order to draw the 
lines. Draw a line for each of the angles in the profile 
A, and also one corresponding to each of the inter- 
mediate points in the curved parts of the stay. Draw 
the line F Gr, representing the oblique cut, and the 
line C D, representing the straight end. Then it will 
be seen that F C D G of Fig. 238, so far as lines are 
concerned, is exactly the same as the molding made 
of clay, shown in Fig. 237. The line F Gr, by the 
definition of a miter, is the ' ' miter line ' ' of this 



Principles of Pattern Cutting. 



75 



molding. It represents the surface against which the 
end of the molding is supposed to fit. Next lay off a 
stretchout of the profile A, in the same manner as de- 
scribed in connection which Fig. 237, all as shown by 
H K in Fig. 238, through the points in which draw 
measuring lines at right angles to it, or, what is the 
same, parallel to the lines of the moldings. Now, 
make each of these lines equal in length to the line of 




in the position shown by the dotted lines in Fig. 238. 
Therefore, instead of using the dividers proceed as 
follows: Place the T-square as shown at E, and, 
bringing it successively against the points in F G, cut 
measuring lines of corresponding number by means of 
a dot or short dash placed across the line. Then a line 
traced as before through the points of intersection thus 
obtained, as shown from L to M, will be the shape of the 
pattern necessary to make it fit against a 
surface placed at the angle represented 
by the miter line F G. By this illus- 
tration it is shown that the T-square 
may be used with great advantage in 
transferring measurements under almost 
all circumstances. Since now the T- 
square is to be used instead of the 
dividers to locate the points in the pat- 
terns, the stretchout line is not needed 
as a starting point from which to meas- 
ure lengths and may, therefore, be 
located at will. For convenience, it 
should be placed as near to the miter 
line as possible. Hence, in practical 



work, supposing 



that the molding rep- 



Fig. S3S.— Obtaining the Envelope of a Molding from a Draiving of the Same by the 

Use of the T-Square. 



corresponding number drawn across the elevation from 
CDtoFG. 

If, as suggested in the previous illustration — that 
is, by using a pair of dividers to measure the length of 
the molding from C D to F G on the several lines — 
these lengths be set off on corresponding lines drawn 
from the stretchout line H K, a pattern will be ob- 
tained in all respects corresponding to the pattern 
shown in Fig. 237, already referred to. By inspection 
of the result thus obtained, however, it will be seen 
that each point in L M is directly under the point of 
corresponding number in line F G, and that the same 
thing may be accomplished by using the T-square placed 



resented by F C D G is not a very 
short piece, the stretchout line, instead 
of being opposite the end C D, would 
be placed somewhere near the line of the p 
blade of the T-square when in its posi- 
tion at E. Should the arm required be 
short, a line drawn opposite the square 
end will serve the double purpose of a 
stretch-out line and of the outline of 
the square end of the pattern. 

By further inspection of Fig. 238, 
it will be seen that, instead of drawing 
the lines from the points in the profile 
A the entire length of the molding, as there shown, 
all that is necessary to the operation is a short line 
corresponding to each of the points of the profile, 
extending only across the miter line F G. The use of 
these lines, it is evident, is only to locate intersections 
upon the miter line. In other words, all that is needed 
is the points in the profile A transferred to the miter 
line F G. The operation of transferring these points 
by short lines, as above described, is termed ' ' drop- 
ping the points " from the profile to the miter line. 

If, instead of the molding terminating against a 
plane surface, as shown by F G in Fig. 238, it be re- 
quired to develop a pattern to fit against an irregular 



76 



The New Metal Worker Pattern Book. 



surface, the method of procedure would be exactly the 
same, simply substituting for the straight line F G a 
representation of that surface. From this it will be 
seen that all that is required to develop the pattern of 
any miter is that a correct representation (elevation or 
plan) of the molding be made, showing the angle of 
the miter, and that a profile be so drawn that it shall 
be in line with the elevation of the molding — its face 




Fig. 239. — Comparison Between a Butt Miter and a Miter Between 
Two Moldings at Any Angle. 

being so placed as to agree with the face of the mold- 
ing — and that points from the subdivisions of the pro- 
file be carried parallel to the molding, their intersec- 
tions with the miter line being marked by short lines. 

In order to more clearly indicate the point desired 
by this summary of requirements, suppose that upon 
each of two pieces of molding made of wood, miters 
at the same angle be cut (right and left) by means of a 
saw, and that they be then placed together, as shown 
in Fig. 239. Now, if a piece of sheet iron, for ex- 
ample, be slipped into the joint, as shown by A, and 
then one arm of the miter be removed what is left will 
be exactly what is shown in Fig. 238. In other words, 
a miter between two straight pieces of molding of the 
same profile is exactly the same as a miter of the same 
mold against a plane, and, hence, the operation of 
cutting the pattern in such a case as shown in Fig. 
239 is identical with that described in Figs. 237 and 
238. 

From this it is plain to be seen that the central 
idea in miter cutting is to bring the points from the 
profile against the miter line, no matter what may be 
its shape or position, and from the miter line into a 
stretch-oi;t prepared to receive them. Inasmuch as all 
moldings, if they do not member or miter with dupli- 
cates of themselves, must either terminate square or 
against some dissimilar profile, it follows that the two 
illustrations given cover in principle the entire cata- 
logue of miters. 

The principles here explained are the funda- 
mental principles in the art of pattern cutting, and 
their application is universal in sheet metal work. It 



would be difficult to compile a complete list of miter 
problems. New combinations of shapes and new con- 
ditions are continually arising. The best that can be 
done, therefore, in a book of this character, is to pre- 
sent a selection of problems calculated to show the 
most common application of principles which, care- 
fully studied, will so familiarize the student with them 
that he will have no difficulty after-ward in working 
out the patterns for whatever shapes may come up in 
his practice, whether they be of those specifically il- 
lustrated or not. 

From the foregoing the following summary of 
requirements, together with a general rule for cutting 
all miters whatsoever, are derived : 

Requirements There must be a plan, elevation 

or other view of the shape, showing the line of the 
joint or surface against which it miters, in line with 
which must be drawn a profile or sectional view of 
same, and this profile must be prepared for use by 
having all its curved portions divided into such a 




Fig. 2Jfl. — Usual Method of Cutting a Square Return Miter. 

number of spaces as is consistent with accuracy and 
convenience. 

It may be remarked here that the division of the 
profile into spaces is only an approximate method of 
obtaining a stretch-out. As theoretically the straight 
distance from one of the assumed points to another 
upon a curved line is less than the distance measured 
around the curve, and the shorter the radius of the 



Princijoles of Pattern Cutting. 



77 



curve the greater is this difference (a chord is less than 
the arc which it subtends,) hence the greater the num- 
ber of points assumed the greater will be the accuracy, 
and a curve of short radius should be divided more 
closely than one of longer radius. The profile thus 
represents practically a succession of plane surfaces. 

Rule. — 1. Place a stretch-out of the profile on a 
line at right angles to the direction of the molding, as 
shown in the plan, elevation or other view, through 

the points in which draw 
measuring lines parallel to 
the molding. 2. Drop lines 
from the points in the profile 
to the miter line or line of 
joint, carrying them in the 
direction of the molding till 
they intersect said line. 3. 
Drop lines from the inter- 
sections thus obtained with 
the miter or joint line on to 
the measuring lines of the 
stretch-out, at right angles 
to the direction of the mold- 
ing. 

In making the applica- 
tion of this rule the student 
must not forget that the 
word profile covers a vast 




Fig. 241. — Comparison Between the Short or Usual Method of Cutting a Square 
Miter and the Method Prescribed by the Bide. 



range of outlines, varying from a simple straight line 
to an entire section of a roof or even more, where large 
curved surfaces are to be treated, and that a rule that 
applies to one can be applied to the others equally 
well. 

The student who gives careful attention to these 
rules will at once remark that the operation of cutting 
a common square miter — that is, a miter between the 
moldings running across two adjacent sides of a square 



building, for example — does not employ a miter line, 
and, therefore, appears to be an exception. Yet it 
has been, remarked that a, thorough understanding of 
how a square miter is cut comprehends within itself 
the science of miter cutting. The square return miter 
— for such is the distinctive name applied to the kind 
of square miter in question — is an exception to the 
general rule only in the sense that it admits of an 
abbreviated method. The short rule for cutting it is 
usually the first thing a pattern cutter learns, and the 
operation is very generally explained to him without 
any reason being given for the several steps taken. 
In many cases it would bother him to cut the pattern 
by any other than the short method, even after he had 
obtained considerable proficiency in his art. Hence it 
is that, to all who have any previous knowledge of 
pattern cutting the rules above set forth seem inade- 
quate, or, to put it otherwise, are a formula to which 
there are exceptions. 

To clear up these doubts in the mind of the stu- 
dent an illustration of the short method of cutting a 
square miter is here introduced, and afterward the long 
method, or the plan which is in strict accordance with 
the rule above given, will be presented, combined with 
the short method, thus showing the relationship and 
correspondence between the two. 

Fig. 240 shows the usual method of developing a 
square return miter, being that in which no plan line 
is employed. The profile A B is divided 
into any convenient number of spaces, as 
indicated by the small figures in the en- 
graving. The stretch-out E F is laid off at 
right angles to the lines of the moldings, 
and, through the points in it, measuring 
lines are drawn parallel to the lines of 
moldings. From the points established in 
the profile lines are dropped cutting corre- 
sponding measuring lines. Then the pattern 
or miter cut G H is obtained by tracing a 
line through these points of intersection. 
In this operation it will be noticed that 
the stipulations of the first part of the rule have beea 
fully complied with — that is, the stretch-out line has 
been drawn at right angles to the lines of the molding, 
and measuring lines have been drawn parallel to those 
lines, but it would seem that the second and third parts 
of the rule as given are not applicable. Apparently no 
miter line has been employed, but the points have 
been dropped directly from the profile into the measur- 
ing lines. 



78 



The New Metal Worker Pattern Book. 



In order to make this clear Fig. 241 is here in- 
troduced in which the proper relation of parts is shown 
and in which the pattern is developed according to 
rule, and in which is also shown the short method and 
how it is derived from the long method. 

As the angle of a return miter can only be shown 
by a plan, the plan becomes the first necessity accord- 
ing to the rule and is shown in the cut by H F K M 
G L, F G showing the line upon which the two arms 
of the molding meet — that is, the miter line. The 
profile A B appears duly in line with one arm of the 
plan II F G L. This arm, then, is the part of which 
the pattern is about to be developed ; accordingly the 
stretch-out line is then drawn at right angles to this 
arm, as shown at C D' ) and the measuring lines drawn 
parallel to the arm. 

The second part of the rule is now carried out ; 
that is, lines are dropped from the points in the profile 
A B to the miter line F G and from thence at right 
angles to F H into the measuring lines, thus obtaining 
the pattern C E'. 

In the upper part of this figure another stretch-out, 
C D, is introduced into which lines have been dropped 
directly from the points in the profile, thus producing 
the pattern at C E, making this part of the figure a 
reduplication of the method employed in the previous 
figure. 

By comparison it will be seen that the two patterns 
C E and C E' are identical. Since the two arms of 
the miter are identical and at right angles to each other, 
the miter line must bisect the angle H F K and be at 
an angle of 45 degrees to either of the two faces H F 
and F K. From this it appears at once that the pro- 
jection of any and all points upon F G from the plan 
line G L toward H is exactly the same as from the 
plan line G M toward K and that the relationship be- 
tween C E and the miter line, and C E' and the miter 
line, is, therefore, the same. Dropping points from 
a profile against a line inclined 45 degrees, as F G, and 
thence on to a stretch-out, gives the same result as 
dropping them on the stretch-out in the first place. 
Hence it is that the portion of the operation shown in 
the lower part of the engraving may be dispensed with. 
This relationship could never occur were the angle of 
the miter anything else than a right angle. 

Another and perhaps simpler explanation of this 
is given in connection with Problem 3, in Section 1 of 
Chapter VI. 

A very common mistake made by beginners in 
attempting to apply the general rule for cutting miters 



as given, is that of getting the miter line in a wrong 
position with reference to the profile. For example, 
instead of drawing a complete plan, as shown by L H 
F K M in Fig. 241, by which the miter line is located 
to a certainty, and in connection with which it is a 
simple matter to correctly place the profile, it is not 
uncommon to attempt the operation by drawing the 
miter line only, placing it either above, below or at 
one side of the profile. The mistake is made by hav- 
ing the line at the side of the profile when it should 
be either above or below it, and vice versa. Fig. 242 
illustrates a case in point. The engraving was made 




Fig. 24%. — -4 Square Face Miter Produced Where a Square Return 
Miter was Intended. 



from the drawing of a person who attempted to cut a 
square return miter by the rule, using a miter line 
only. By ]}lacing the miter line E F at the side in- 
stead of below the profile, a square face miter- — such 
as would be used in the molding running around a 
panel or a picture frame — was produced in place of 
what was desired. 

In order to avoid such errors the reader is recom- 
mended to a careful perusal of the chapter on Linear 
Drawing (Chapter III), where the relation existing be- 
tween plans, elevations and sections or profiles is thor- 
ougly explained. It is better to draw a complete plan, 
as shown in Fig. 241, thus demonstrating to a cer- 



Principles of Pattern Cutting. 



79 



tainty the correct relationship of the parts, than to 
save a little labor and run the risk of error. 

As remarked in the earlier part of this chapter, 
some labor is often necessary before the requirements 
mentioned above in connection with the rule can be 
fulfilled. Sometimes a miter line must first be de- 
veloped, and sometimes the profile of a molding must 
undergo a change of profile known as raking. It is 
believed that the principles underlying these opera- 
tions are made sufficiently clear in connection with the 
problems in which they are involved not to need es- 
pecial explanation in this connection. Suffice it to 
say that, in many instances, half the work is done in 
the getting ready. 

FormSo 



Regiaflar Taperta 

(FLARING WORK.) 

This subject embraces a large variety of forms of 
frequent occurrence in sheet metal work, and the de- 
velopment of their surfaces comes under an altogether 
different set of rules than those applied to parallel 
forms. 

Before entering into the details of these methods 
it will be best to first define accurately what is here 
included by the use of the term. These forms include 
only such solid figures as have for a base the circle or 
any of the regular polygons, as the square, triangle, 





Fig. 24S. — A Right Cone Generated by the Revolution of a Right- 
Angled Triangle about its Perpendicular. 

hexagon, etc. ; also figures though of unequal sides 
that can be inscribed within a circle, and all of which 
terminate in an apex located dh - ectly over the center 
of the base. 

While the treatment of these forms has been said 
to be altogether different from that of parallel forms 
(.here are some points of similarity to which the stu- 
dent's attention is called that may serve to fix the 
methods of work in his memory. 



Whereas in parallel forms the distances of the 
various points in a miter are measured from a straight 
line drawn through the mold near the miter for that 
purpose, as G D, Fig. 238, the distances of all 
points in the surfaces of tapering solids produced by 
the intersection of some other surface are measured 
from the apex upon lines radiating therefrom; and 




Fig. 2U-—A Right Gone with Thread Fastened at the Apex to which 
are Attached Points Marking the Upper and Lower Bases of a 
Frustum. 

whereas the distance across parallel forms (the stretch- 
out) is measured upon the profile, the distance across 
tapering forms is measured upon the perimeter of the 
base. 

Patterns are more frequently required for por- 
tions of frustums of these figures than for the com- 
plete figures themselves and the methods of obtaining 
the pattern of coverings of said frustums is simply to 
develop the surface of the entire cone or pyramid and 
by a system of measurements take out such parts as 
are required. 

As the apex of a cone is situated in a perpendic- 
ular line erected upon the center of its base, it must 
of necessity be equidistant from all points in the cir- 
cumference of the base. 

In works upon solid geometry the cone is de- 
scribed as a solid generated by the revolution of a 
right-angle triangle about its vertical side as an axis. 
This operation is illustrated in Fig. 243, in which it 
will be seen that the base E D of the triangle C E D 
is the radius which generates the circle forming the 
base of the cone, and that the hypothenuse C D in 
like manner generates its covering or envelope. 



80 



The New Metal Worker Pattern Book. 



If a plane be passed through a cone parallel to 
the base and at some distance above it, the line which 
it produces by cutting the surface of the cone must 
also be a circle, because it, like the base, is perpen- 
dicular to the axis. The portion cut away is simply 
another perfect cone of less dimensions than the first, 




Fig. 245.— Frustum of a Right Cone, the Dotted Lines Shoiving the 
Portion of the Cone Removed to Produce the Frustum. 

while the. portion remaining is called a frustum of a 
cone. AFC, Fig. 244, is a cone, and BDEC, Fig. 
245, is a frustum. The line B E, Fig. 244, shows 
where the cone is cut to produce the frustum. 

If, having a solid cone of any convenient material, 
as wood, a pin be fastened at the apex C of the same, 
as shown in Fig. 244, and a piece of thread be tied 




Fig. 246.— Envelope of the Cone and Frustum Described by the Pin 
and Thread in Fig. 244. 

thereto, to which are fastened points B and A, corre- 
sponding in distance from the apex to the upper and 
lower bases of the frustum, and the thread, being drawn 
straight, be passed around the cone close to its surface, 
the points upon the thread will follow the lines of the 
bases of the frustum throughout its course. If then, 
taking the thread and pin from the cone, and fastening 



the pin as a center upon a sheet of paper, as shown in 
Fig. 246, the thread be carried around the pin, keep- 
ing it stretched all the time, the track of the points 
fastened to the thread will describe upon the paper the 
shape of the envelope of the frustum, as shown by 
Gr D E F. By omitting the line produced by the 
upper of the two points, the envelope of the complete 
cone Gr C F will be described. The length of the arc 
Gr F described by the point A attached to the thread 
may be determined by measuring the circumference of 
the base of the cone by any means most available. 
The usual method is to take between the points of the 
dividers a small space and step around the circumfer- 
ence of the circle of the base and set off upon the circle 
of the pattern the same number of spaces. 




Fig. 247. — Unfolding the Envelope of a Right Cone. 

The development of the envelope of a cone may 
be further illustrated by supposing that, in the case 
of the wooden model, it be laid upon its side upon a 
sheet of paper and rolled along until it has made one 
complete revolution ; a point having been previously 
marked upon the line of its base by which to deter- 
mine the same. The base B, Fig. 247, thus becomes 
stretched out as it were, describing the line C D upon 
the paper, while the apex A, having no circumfer- 
ference, remains stationary at the point A 1 . The lines 
G A' and D A 1 represent the contact of the side of the 
cone at the beginning and at the finish of one revolu- 
tion. 

As in the case of dividing the profile in parallel 
forms, this method is, theoretically, only approximate 
in accuracy, but the difference is so slight practically 
that it is not worth considering. Of course, the shorter 



Principles of Pattern Cutting. 



81 



the spaces are the greater is the accuracy. This 
method has, however, another significance which will 
be pointed out later on, which will help to simplify 
the solution of all tapering forms. 




Fig. 243. — A Cone Truncated Obliquely. 

If it is required that the cone should be truncated 
obliquely, as shown in Fig. 248, it will be seen that 
all the points in the upper line of the frustum are at 
different distances from the base, or, what amounts to 
the same thing, from the apex of the original cone, 
hence some method of measuring these distances must 
be devised. 

To explain the principles here involved more 




Fig. 249.— Plan and Elevation from which to Construct a Wooden 
Cone for Purposes of Illustration. 

clearly, suppose that a cone be cut from a solid block of 
wood and of a hight and width to agree with some par- 
ticular drawing, as, for instance, the one shown in Fig. 
249. Divide the circle of the base E F upon the drawing 
into a convenient number of parts or spaces and mark 



the same number of points and spaces upon the edge 
of the base of the wooden cone, and from each of these 
points draw upon the sides of the wooden cone straight 
lines running to its apex. 

A correct elevation of these lines upon the draw- 
ing may be obtained by carrying lines from the divisions 
or points in the plan of the base vertically till they 
strike the line of the base B C in the elevation, as shown 
in Fig. 250, thence to the apex A, cutting the line 
G H. 

Now, if by means of a saw the upper part of the 
wooden cone be removed, being cut to the required 
angle as shown by the oblique line G II in the draw- 
ing, an opportunity is given, by the lines upon the 




Fig. 250. — Method of Obtaining the Lines upon the Elevation. 

part of the cone cut away, of measuring accurately the 
distance of each point of the curve thus produced from 
the apex. 

Then as all points in the base B C are equidistant 
from the apex A, to lay out the pattern of this frustum, 
first describe an arc of a circle whose radius is equal to 
the length of the side (or slant hight) of the cone A B, 
Fig. 250. Make this arc in length equal to the cir- 
cumference of the base B C of the cone by means of 
the points, as previously described. To avoid con- 
fusion number these points 1, 2, 3, etc., from the start- 
ing point B, and from each of these points draw lines 
to the center of the arc, all as shown in Fig. 251. 

Now, replacing that portion of the cone which 
was cut away so as to identify the lines upon its sides 



The New Metal Worker Pattern Book. 



by the numbers at the base, the length of each line 
from the apex down to the cut can be measured by the 
dividers and transferred to the lines of the same num- 
bers in the diagram, Fig. 251, as shown between G 
and H. 

All this no doubt is quite simple when the model 
is at hand upon which to make the measurements. It 
is quite evident that it will not do to measure the dis- 
tance upon the drawing, Fig. 250, from the apex A to 
the points of intersection on the line G H because the 
sides of the cone having an equal slant of flare all 
around, the lines upon the drawing do not represent 
the real distances except in the case of the two outside 
lines ; the slant hight of a cone or any part of a cone 
being greater than the vertical hight of same part. 
But as these two outside lines do represent the correct 




Fig. 251.— Method of Deriving the Pattern of a Frustum from the 
Wooden Model. 

slant of the cone on all sides, either one of them may 
be taken as a correct line upon which to measure these 
distances; that is, as a vertical section through the 
cone upon any or all of the lines drawn upon its sides. 
To make it a perfect section upon any one of 
these lines, say line 5, it is simply required that the 
position of the point of intersection of line 5 with the 
line G H be shown, which is done by carrying this 
point horizontally across till it strikes the side of the 
cone A B at 5, as illustrated in Fig. 250. The result 
of repeating this operation upon all the other lines is 
as though a thread or wire were stretched from the 
apex down along the side of the cone to the point B 
in the base and the cone were turned upon its axis, 
and as each line upon the side passes under the thread, 
the point where it cuts the intersecting plane G H 
where marked thereon, thus collecting all the points 
into one section as it were. 



This operation is fully shown in Fig. 252, to 
which is added the development of the pattern, which 
is exactly the same as that shown in Fig. 251, the dis- 
tances of the points between G and H from A being 
obtained in this case from the points upon the line 
A B, instead of from the model, as before. The points 
on A B are transferred to lines of corresponding num- 
ber in the pattern by means of the compasses, as 
shown. 

Should the frustum of which a pattern is required 




Fig. 252.— Method of Deriving the Pattern from the Drawing. 

have both its upper and lower faces oblique to the 
axis of the cone a level base can be assumed at a con- 
venient distance below the lower face of the frus- 
tum from which the circumference can be obtained 
and then both the upper and lower faces of the frus- 
tum can be developed by the method just described. 

A right cone having an elliptical base might seem 
to belong in the same class with regular tapering forms, 
but as the distance from its apex to the various points 
in the perimeter of its base is constantly varying, it is 
therefore placed in the class with irregular forms in 



Principles of Pattern Cutting. 



83 



the following section of this chapter, where it will be 
duly discussed. But as tapering articles of elliptical 
shape are of frequent occurrence, and as the circle is 
much easier made use of than the ellipse, siich articles 
are usually designed with approximate ellipses com- 
posed of arcs of circles. This method is in many cases 
especially desirable, as articles so designed have an 
equal amount of flare or taper on all sides, which 
would not be the case if they were cut from elliptical 
cones. It will thus be seen that an article designed in 
that manner is the envelope of a solid composed of as 
many portions of frustums of right cones as there were 
arcs of circles used in drawing its plan. 

In Fig. 253 is shown the usual method of draw- 
ing the plan and elevation of an elliptical flaring ar- 
ticle, the outer curve of the plan ACBD being the 
shape at M N of the elevation, while the inner curve 
I G V J is the plan at the top K L. As many 
centers may be employed in drawing the curves of the 
plan of such an article as desired, all of which is ex- 
plained in the chapter on Geometrical Problems (Chap. 
TV.), Problems 73, 76 and 78. To simplify matters only 
two sets of centers have been employed in the present 
drawing, all as indicated by the dotted lines drawn from 
the various centers and separating the different arcs of 
circles. Reference to the plan now shows that that 
portion of the article included between the points E 




Fig. 263. — Usual Method of Draiuing an Elliptical Flaring Article 

C W V G H is a position of the envelope of a cone 
the radius of whose base is D C and whose apex is 
located at a point somewhere above D ; and likewise 
that that portion included between the points X A E 
H I J is part of a cone the radius of whose base is 
A F and whose apex is somewhere upon a line erected 
at F. Thus four sectors cut from cones of two differ- 
ent sizes go to make up the entire solid of which the 
article shown in Fig. 253 is a frustum. 



To determine the dimensions, then, of such cones 
it is necessary to construct a diagram such as that 
shown in Fig. 254, which is in reality a section upon 
the line ED of the plan, in which P and P R 
are respectively equal to E D and E F of the plan. 
At points and R, Fig. 254, erect perpendicular lines 
J and R Z indefinitely. Upon J set off S equal 
to the straight hight O K of the frustum, Fig. 253, and 




Fig. 254. — Diagram Constructed to Determine the Dimensioyis of the 
Cones, Portions of which are Combined to Make up the Article 
Shown in Fig. 253. 

draw S U parallel to P, which make equal in length 
toDH. A line drawn through the points P and U 
will then represent the slant or taper of the frustum, 
as shown at M K of the elevation, and if continued till 
it intersects with the perpendiculars from O and P will 
determine the respective hights of the two cones, as 
shown by Z and J. Then P JO is the triangle which, 
if revolved about its vertical side J 0, will generate 
the cone from which so much of the figure as is struck 
from the centers G and D in Fig. 253 is cut; and P Z R 
is the triangle which if revolved about its vertical side 
Z R will generate the cone from which the end pieces 
of the article are taken. To present this before the 
reader in a more forcible manner, several pictorial illus- 
trations are here introduced in which the foregoing 
operations are more clearly shown. In Fig. 255 is 
shown a view of the plan of the base A C B D of Fig. 
253 in perspective, in which the reference letters are 
the same as at corresponding parts of that plan, and 
upon which is represented, in its correct position, a 
I sector of the larger cone from which the side portions 



84 



The New Metal Worker Pattern Book. 



of the frustum are taken. Thus the triangular sur- 
faces FDE and F D W, being sections of the cone 
through its axis, correspond to the triangle J P of 
the diagram, Fig. 254. In Fig. 256 two additional 
sectors from the smaller cone previously referred to 
are represented as standing upon the adjacent portions 
of the plan from which their dimensions were derived. 
Thus C F and H D, the center lines of their bases, 
correspond respectively to A F and Y B of the plan, 
Fig. 253, and the triangles L F Gr and M II K, being 
radial sections of the cones, correspond with the tri- 
angle ZEPof the diagram. In Fig. 257 is presented 
the opposite view of the combination seen in Fig. 256, 




Fig. S55. — Perspective View of the Plan in Fig. 253, with a Sector of 
the Larger Cone in Position. 

showing at C B and D A the joining of their outer 
surfaces or envelopes. 

As previously remarked, two sets of centers only 
were employed in constructing the plan, Fig. 253, for 
the sake of simplicity. Had a third set of centers 
been made use of the arrangement of sectors of cones 
shown in Figs. 256 and 257 would have been supple- 
mented by a pair of sectors, cut from a cone of inter- 
mediate size, which would have been placed on either 
side of the large sector between it and the smaller 
ones, all being joined together upon the same general 
principle as before explained. Beference to Fig. 220 
in the chapter on Geometrical Problems shows at J L 
P, P S W and W TJ V what the relative position of 
their bases would be. If it be desired to complete the 
solid, which has been begun in Fig. 256, it will first be 
necessary to cut away that portion of the middle sector 



which stands over the space F H B, Fig. 256. Such a 
cut might be begun upon the line F H, and passing 
vertically through the points L and M would finish 
through the curved surface of the further or curved 
side of the sector. The cut thus made between the 
points L and M is shown at D C in the other view, 
Fig. 257, and is by virtue of the conditions a hyper- 
bola. (See Def. 113, Chap. I.) The piece necessary 
to complete the solid would then be a duplicate of the 
shape remaining after making the above described cut, 
the outer surface of which is shown by A B C D of 
Fig. 257. The complete solid would then have the 
appearance shown in Fig. 258. 




Fig. 256. — The Same Plan Showing Ttvo Sectors of the Smaller Cone 
in Position Joining the Larger One. 

By thus resolving the solid from which the ordi- 
nary elliptical flaring article is cut into its component 
elements the process of developing its pattern may be 
more readily understood. This process may now be 
easily explained by returning to the string and pin 
method which was made use of in connection with the 
simple cone in the earlier part of this section. 

In Fig. 257 is shown a line some distance above 
the base representing the top of the frustum shown by 
K L in the original elevation, Fig. 253. It also 
shows a pin fastened at the apex of the middle conical 
sector to which is attached a thread carrying points 
G- and H representing the irpper and lower surfaces of 
the frustum. Now, if the string be drawn tight and 
passed along the side of the larger sector of the cone 
from A to B the points will follow the upper and 
lower bases of the frustum. When the point B is 



Principles of Pattern Cutting. 



85 



reached, if the finger be placed upon the thread at the 
apex of the lesser cone, shown at C, and the progress 
of the thread be continued, the points will still follow 
the lines of the bases of the frustum. If the pin and 




Fig. 257. — Opposite View of Parts Shown in Fig. 256, with String 
Attached to a Pin at the Apex of the Larger Sector, 

thread be taken from the cone and transferred to a 
sheet of paper, as shown in Fig. 259, the pin A being 
used as a center and the thread as a radius, the points 
will describe the envelope of the frustum. First, the 
radius is used full length, as shown by A L K, and 
arcs L M and K H are drawn in length respectively 




Fig. S5S. — The Completed Solid of which the Ordinary Ellipitical 
Flaring Article is a Part. 

equal to their representatives H G V and E C W of 
the original plan, Fig. 253. Then a second pin is 
put through the string, as shown at B, thus reducing 
the radius to the length of the side of the lesser cone, 
and arcs are struck in continuation of those first de- 



scribed, making the length of the additional arcs 
equal to those of their corresponding arcs HI J and 
E A X of the original plan. 

As the lengths of the sides of the larger and 




Fig. 259. — The Pin and Thread taken from Fig. 257 and Used in 
Describing the Envelope. 

smaller cones above made use of are by construction 
equal to J P and Z P, the hypothenuses of the tri- 
angles, Fig. 254, by whose revolution they were 
generated, those distances may therefore be taken at 
once from that diagram by means of the compasses 
and used as shown in Fig. 259. 

Reference has been made above to the difference 
between the circumference of the circle of the base 
obtained by means of the points and spaces (which 
method becomes a necessity to the pattern cutter) and 



-An Arc Compared with its Chord. 




An explanation of this differ- 
next class of regular tapering 



the real circumference, 
ence will lead to the 
figures — viz. : pyramids. 

In the accompanying diagram, Fig. 260, ABC 
represents the arc of a circle of which the straight line 
A C is the chord, being the shortest distance between 
the two points A and C. Therefore, when dividing 
a circle by means of points for purposes of measure- 
ment, the pattern cutter is in reality using a number 
of chords instead of the arcs which they subtend. 



86 



The New Metal Worker Pattern Book. 



In the practice of obtaining the circumference or 
stretch-out of a circle the space assumed as the unit of 
measure should be so small that there is no perceptible 
curve between the points and, of course, no practical 
difference between the length of the chord and the 
length of the arc. 

It will thus be seen that the circle representing 
the base of a cone has in reality become in the hands 
of the pattern cutter a many sided polygon and that 
the cone is to him a many sided pyramid. As one of 
the conditions in describing a regular polygon is that 
its angles shall all lie in the same circle, so the angles 
or hips of a pyramid must lie in the surface of the cone 
whose base circumscribes the base of the pyramid and 
whose apex coincides with the apex of the pyramid. 
Viewed in this light then, the lines which were drawn 
upon the outside of the wooden cone for the purpose 
of measurement in the illustration used above become 
the angles or hips of a pyramid and may be used for 
that puipose in exactly the same manner. 

In developing the pattern of a frustum of a cone 
the line connecting the points between Gr and H, Fig. 
251, is supposed, of course, to be a curved line, while 
in the case of a pyramid (the points or angles of the 
pyramid being further apart and the sides of a pyramid 
being flat instead of curved) the lines of the pattern 
connecting the points would be straight from point to 
point. 

Irregmlar Forms 

(TRIANGULATION.) 

In some classes of sheet metal work certain forms 
arise for which patterns are required, but which cannot 
be classified under either of the two previous sub- 
divisions. Their surfaces do not seem to be generated 
by any regular method. They are so formed that 
although perfectly straight lines can be drawn upon 
them (that is, lines running parallel with the form), 
such straight lines when drawn would not be parallel 
with each other ; neither would they slant toward each 
other with any degree of regularity. 

While in the systems described in the two previous 
portions of this chapter distances between lines running 
with the form measured at one end of an article govern 
those at the other end, in the forms considered in this 
department these distances are continually varying and 
bear no such relation to each other. Thus in parallel 
forms (moldings) the distance between any two lines 
running with the form is the same at both ends of the 



article, while in conical shapes all lines running with 
the form tend toward a common center or vertex, so 
that the distances between such lines at one end of the 
article (provided it does not reach to the vertex) bear 
a regular proportion to the distances between them at 
the other end. Hence, in the development of the 
pattern of an irregular form it becomes necessary to 
drop all previously described systems and simply pro- 
ceed to measure up its surfaces, portion by portion, 
adding one portion to another till the entire surface 
has been covered. 

To accomplish this end one of the most simple 
of all geometrical problems is made use of. to which 
the reader is referred (Chap. IV., Problem 36) — viz. : 
To construct a triangle, the lengths of the three sides being 
given. 

As from any three given dimensions only one 
triangle can be constructed, this furnishes a correct 
means of measurement ; and the solution of this prob- 
lem in connection with a regular order and method of 
obtaining the lengths of the sides of the necessary tri- 
angles constitutes the entire system. To carry out 
this system it simply becomes necessary to divide the 
surface of any irregular object into triangles, ascertain 
the lengths of their sides from the drawing, and re- 
produce them in regular order in the pattern, and hence 
the term Triangulation is most fittingly applied to 
this method of development of surfaces. 

In all articles whose sides lie in a vertical plane, 
distances can be measured in any direction across their 
sides upon an elevation of the article, but when the 
sides become rounded and slanting the length of a 
line running parallel with the form cannot be measured 
either upon the elevation or the plan. The elevation 
sives the distance from one end of the line to the other 
vertically or as it appears to slant to the right or left, 
but the distance of one end of the line forward or back 
of the other can only be obtained from the plan which 
while supplying this dimension does not give the hight. 
Consequently the true length of any straight line lying 
in the surface of any irregular form can only be ascer- 
tained by the construction of a right-angle triangle 
whose base is equal to the horizontal distance between 
the required points, and whose altitude is equal to the 
vertical distance of one point above the other, the 
hypothenuse giving the true distance between the 
points, or, in other words, the required length of the 
line. 

For illustration, Fig. 261 shows an article which 
may be called a transition piece, the base of which, 



Principles of Pattern Cutting. 



S7 



A B C D of the plan, is a perfect circle lying in a 
horizontal plane, E H of the elevation. Its upper sur- 
face, however, IST P Q of the plan, is elliptical in 
shape and besides being placed at one side of the 
center is also in an inclined position, as shown by F G 
of the elevation. To the right of this plan is another 
drawing of the same, A' B' C D', turned one-quarter 
around from which, and the elevation, is projected 
another view, J' K' L M, which may be called the 



perimeter at the top. As F G, the distance across the 
top, is greater than N P (its apparent width in the 
plan), the curve 1ST P Q evidently does not give the 
correct distance around the top, and therefore a correct 
view of the top must be obtained. The method of 
accomplishing this does not differ from many similar 
operations described in connection with parallel forms 
and is clearly shown in the drawing. Considering 
N P Q as a correct plan or horizontal projection of 



Ml 



v/ / / l / y 

/TRUE PROFILE/ 
/ /OF TOP / ] H 

ijrhh-Li //}' 



//// /I 



/" l ''Hi 
// / / 1/ 



3 i 



5 9(07 
10 8 
DIAGRAM OF 
DOTTED LINES 




IOC 




Fig. 262. Fig. 261. 

Plans, Elevations, etc., of an Irregular Shaped Article, Illustrating the Principles of Triangulation. 



front and which will assist in obtaining a more perfect 
conception of the shape of the article. A comparison 
of the three views shows that the slant of the sides is 
different at every point, and that the only dimensions 
of the article which can be measured directly upon the 
drawing are the circumference of the base and the 
slant hight, as given at E F, H G and L M. 

Before a pattern of its side can be developed it 
will be necessary to ascertain its width (or distance 
from base to top) at frequent intervals and also its 



the top, one-quarter of it, as !N", may be divided by 
any convenient number of points and their distances 
from N P set off upon the parallel lines drawn from 
N" P", thus obtaining 0" N", one-quarter of the cor- 
rect curve. It is more likely, however, that the cor- 
rect shape of the top N" 0" P" would be given, from 
which it would be necessary to obtain its correct 
appearance in the plan, which would be accomplished 
by drawing the normal curve in its correct relation to 
the line F G, as shown by 1ST" O" P", when the raking 



88 



The New Metal Worker Pattern Book. 



process could be reversed, thereby developing the 
curve N P one-half of the plan of top. 

Preparatory to obtaining the varying width of the 
pattern of the side, a number of points must be fixed 
upon in the curves of both top and bottom from which 
to take the measurements. As one-quarter of the top 
is already divided into spaces, another quarter, P, 
may be divided into the same number of spaces (also 
dividing O" P" into the same space as 0" N"). If 
N" 0" P" is the normal curve of the top it would very 
naturally be divided into equal spaces by the dividers, 
as is usual in such cases, while the spacing in N P 
would be the result of the operation of raking. It is 
advisable to have the spaces in N"0"P" all equal to 
each other, as it is from this curve that the stretch-out 
of the top of the pattern is to be derived, the conven- 
ience of which will become apparent when the pattern 
is developed. 

The quarter P is used in connection with the quar- 
ter N, because these two combined constitute a half of 
the top curve lying on one side of the line A C of 
the plan which divides the article into symmetrical 
halves, it being only necessary, when the shape of an 
article permits, to obtain the pattern of one-half and 
then to duplicate by any convenient means to obtain the 
other half. 

The corresponding half of the plan of the base, 
therefore, A B C, must also be divided into the same 
number of equal spaces as were used at the top, all as 
shown in the drawing, and both sets of points should 
be numbered alike, beginning at the same side. 

Having thus fixed the points from which measure- 
ments across the pattern of the side are to be taken, 
next draw lines across the plan connecting points of 
like number, as shown by the full lines in the plan. 
This divides the entire side of the article into a num- 
ber of four-sided figures ; but as it is necessary, as 
shown above, to have it divided into triangles, each 
four-sided figure may now be redivided by a line drawn 
through its opposite angles, thus cutting it into two 
triangles. In other words, each point in the base 
should be connected with a point of the next lower 
'number (or higher, according to circumstances) in the 
curve of the top, and these lines should be dotted in- 
stead of full lines for the sake of distinction and to 
avoid confusion in subsequent parts of the work. 
Thus 1 of the base is connected by a dotted line with 
0' of the top, 2 of the base with 1' of the top, etc. 

In respect to which is the best way to run the 
dotted lines, common sense will be the best guide. 



Thus, in the space bounded by the lines 4 4' and 5 5', 
it is plainly to be seen that there would be greater ad- 
vantage and less liability of error in connecting 5 of the 
bottom curve with 4' of the top than in crossing the 
line from 4 of the bottom to 5' of the top, for the rea- 
son that in the former case the triangles produced 
would be less scalene or acute. 

The next step is to devise a means of determining 
the true lengths which these lines represent or, in other 
words, their real length as they could be measured if a 
full size model of the article were cut from a block of 
wood or clay iqjon which these lines had been marked, 
as shown upon the drawing. 

The lines upon the plan, of course, only show the 
horizontal distances between the points which they 
connect. The vertical night above the base of any of 
the points in the upper curve can easily be found by 
measuring from its position upon the line F G of the 
elevation perpendicularly to the base E H. Therefore, 
having both the vertical and the horizontal distance 
given between any two points, it is only necessary to 
construct with these dimensions a right angle triangle, 
and the hypothenuse will give their true distance apart. 
Thus in Fig. 262 ah is equal to the line 4 4' of the 
plan, while a c is made equal to 4 4' of the elevation. 
Consequently c h represents the true distance between 
the points 4' of the top and 4 of the base. Therefore, 
to obtain alb of these hypothenuses in the simplest 
possible manner, it will be necessary to construct one 
or two diagrams of triangles. To avoid confusion it 
is better to make two ; one for obtaining the distances 
represented by the full or solid lines drawn across the 
plan and the other for those of the clotted lines. To 
do this extend the base line EH of the elevation, as 
shown at the left, at any convenient points, in which, 
as B. and S, erect two perpendicular lines. Project 
lines horizontally from all the points in F G, cutting 
these two lines as shown, and number the points of 
intersection. (Some of the figures are omitted in the 
drawing for lack of sj)ace.) From E set off on the base 
line distances equal to the lengths of the solid 
lines of the plan 1 1', 2 2', 3 3', etc., numbering the 
points 1, 2, 3, etc., as shown, and connect jioints of 
similar number upon the base with those upon the 
perpendicular. From S set off on the base line dis- 
tances equal to the lengths of the dotted lines of the 
plan 1 0', 2 1', 3 2', etc., and number them to corre- 
spond with figure upon the line of the base ABC. 
Thus make 1 S equal to 1 0' of the plan, 2 S equal to 
2 1' of the plan, 3 S equal to 3 2', etc., and connect 



Principles of Pattern Cutting. 



89 



each point in the base with the point of next lower 
number upon the perpendicular by a dotted line, as 1 
on the base with on the perpendicular, 2 with 1, 3 
with 2, etc. The entire surface of the piece for which 




Fig. 26$.— Top, Back and Bottom for a Model of One-half the 
Article Shoivn in Fig. 261. 

a pattern is required has thus been cut up into two sets 
of triangles, one set having the spaces upon the base 
line ABC, which are all equal, for their bases, and 
the other set having the spaces in the curve N" 0" P" 
of the top, also equal to each other, as their bases, and 
each separate triangle having one solid line and one 
dotted line as its sides. 

In all of this work the student's powers of mental 
conception are called into play. The shape of the sur- 
face, which is yet to be developed, has been spoken of 
as if it ideally existed— in fact, it must exist in the mind 
or imagination of the operator in order to make him 
intelligent as to what he is doing. If this fails him, 
he can resort to a model which can easily be con- 
structed (full size or to scale, according to convenience) 
as follows : Describe upon a piece of cardboard or 
metal the shape E F Gr H, Fig. 2G1, to which add on 
its lower side, E H, one-half of the plan of the bottom, 
ABC, with the curve NOP and the solid lines con- 
necting it with the outside curve traced thereon. Also 
add on its upper side, F G, one-half the shape of top, 
N" 0" P", marking the points 1, 2, 3, etc., upon its 
edge. Now cut out the entire shape in one piece, as 
shown in Fig. 263, and bend the same at right angles, 
on the lines F Gr and E H. Small triangles of the 
shape and size of each of the triangles shown in the 
diagram of solid lines, Fig. 261, as B, 1 1 E, 2 2E, 



etc., can now be cut out and placed upon the portion 
representing the bottom, each with its base upon the 
solid line which it represents, at the same bringing the 
apex of each to the corresponding number on the top. 
These can be fastened in place by bits of sealing 
wax, or if cut from metal the whole can be soldered 
together. 

The hypothenuses of the various triangles will 
thus represent the true distances across the pattern 
upon the solid lines of the plan, while the distances 
upon the dotted lines can be represented by pieces of 
thread or wire, placed so that each will reach from the 
point at the base of one of the triangles to the point at 
the top of the one next it. If constructed of metal 
two or three triangles will suffice to give the model 
sufficient rigidity, and the remaining points can be 
connected by pieces of wire, using a different kind of 
wire to represent the distances on the dotted lines. 

In Fig. 261, is shown a pictorial representation of 
a model constructed, as above described, from the draw- 
ings shown in Fig. 261. In the illustration the tri- 
angles 2, 5 and S only are shown in position, their 
hypothenuses connecting points of similar number in 
the upper and lower bases. The other points are rep- 
resented as being connected by wires or threads repre- 
senting both the solid and the dotted hypothenuses in 
the diagrams of triangles in Fig. 261. Such a model if 
constructed will give a general idea of the shape of the 
entire covering, and at the same time of the small 
pieces, or triangles, of which the covering is composed, 
with all the dimensions of each. If all of the spaces 
formed upon this skeleton surface could be filled in 




Fig. 264— Perspective View of Cardboard Model of One-half the 
Article Shown in Fig. 261. 

with pieces of cardboard or metal just the size of each 
and the whole removed together and flattened out (each 
piece being fastened to its neighbor at the sides), it 
would constitute the required pattern, the same as will 
be subsequently obtained by measurements taken from 
the drawing, and as shown in Fig. 265. 



90 



The New Metal Worker Pattern Book. 



Having by means of the diagrams of triangles in 
Fig. 261 obtained the lengths of all the sides it is now- 
only necessary to construct successively each triangle 
in the manner described in Chapter IV, Problem 36, 
remembering that the last long side of each triangle 
used is also the first long side of the next one to be 
constructed. Therefore, at any convenient place draw 
any straight line, A N of Fig. 265, which make equal 
to the real distance from A to N, Fig. 261, which has 
been found to be the distance of the diagram of 
solid lines. To conduct this operation with the great- 
est economy and ease it is necessary to have two pairs 
of dividers, which shall remain set, one to the spaces 
upon the plan of the base ABC, and the other to the 
spaces upon N" 0" P", and a third pair for use in 
taking varying measurements. From A of Fig. 265 
as a center, with a radius equal to 1 of the plan, Fig. 
261, describe a small arc, and from N as a center, with 
a radius equal to the true distance from N to 1 of the 
plan, which has been found to be 1 of the diagram 
of dotted lines, describe another arc, cutting the first 
one as shown at the point 1, Fig. 265. The triangle 
thus constructed represents the true dimensions of one 
indicated by the same figures of the plan. Next from 
N of the pattern as a center, with a radius equal to 
N" 1 of true profile of top, Fig. 261, describe a small 
arc, which cut with one struck from point 1 of pat- 
tern as a center, with a radius equal to 1 1 of the dia- 
gram of solid lines, thus locating point 1' of pattern. 
This triangle is, in turn, succeeded by another whose 
sides are next in numerical order, that is 1 2 of the base 
and 1 2 of the diagram of dotted lines. Thus the 
operation is continued, always letting the spaces of the 
circumference of base succeed one another at one side 
of the pattern, and the spaces upon the true profile of 
top succeed one another at the other side of the pat- 
tern, until all the triangles have been laid out as 
shown by AN P C, Fig. 265, which will complete 
one-half the entire pattern. 

It is not necessary to draw all of the dotted or 
solid lines across the pattern, as the points where the 
small arcs intersect are all that are really needed in 
obtaining the outlines of the pattern, but it is often 
advisable to draw them as well as to number each new 
point as obtained, in order to avoid confusion and 
insure the order of succession. 

In dividing the curves of top and bottom into 
spaces, such a number of points should be taken as 
will insure the greatest accuracy, as in the case of 
dividing a profile. Thus too few would give too short 



a stretch-out, while if the spaces were too small error in 
transferring their lengths might result, which would be 
increased as many times as there were spaces. 

Under the head of transition pieces may be in- 
cluded a large number of forms having various shaped 
polygonal or curved figures as their upper and lower 
surfaces, placed at various angles to each other, some- 
times centrally located as they appear upon the plan 
and sometimes otherwise. It often happens that one 
surface or termination is entirely outside the other in 
that view, forming an offset between pipes of differing 
sizes and shapes. Sometimes such an offset takes a 
curved form, constituting a curved elbow of varying 
section throughout its length, in which case it consists 
of a number of pieces, each with a different shape at 
either end. With such forms may be classed the ship 




Fig. 265.— One-half Pattern of Side of Article Shown in Fig. 281. 

ventilator, whose lower end is usually round and hori- 
zontal and whose upper end is enlarged and elliptical 
and stands in a vertical position, the whole being com- 
posed of five or six pieces. In such cases, when the 
shape and position of the two terminating surfaces only 
are given, it becomes necessary to assume or draw as 
many intermediate surfaces as there are joints required, 
each of such a shape that the whole series will form a 
suitable transition between two extreme shapes. It 
may be remarked, that what have been spoken of 
here as "surfaces" do not necessarily mean surfaces 
of metal forming solid ends to the pieces described, 
but simply outlines upon paper to work to, as more 
often the "surface" is really an opening. 

Still another class of forms demanding treatment 
by triangulation result from the construction of arches 



Principles of Pattern Cutting. 



91 



cut through curved walls, as when an arch of either 
round or elliptical form, as a door or window head, is 
placed in a circular wall in such a manner that its sides 
or jambs are radial, or tend toward the center of the 
curve of the wall. It will be seen that the soffit of 
such an arch is similar in shape to the sides of a transi- 
tion piece, having what might be called its upper and 
lower surfaces curved and placed vertically. In such 
cases it is best to consider the horizontal plane passing 
through the springing line of the arch as the base from 
which to measure the hights of all points assumed in 
the outer and inner curves. 

It is believed that a sufficient number of this gen- 




7 P 
Fig. 266. — Elevations and Plan of an Elliptical Cone. 

eral class of problems will be found in the third section 
of the chapter on Pattern Problems to enable the care- 
ful student to apply the principles here explained to 
any new forms that might present themselves for his 
consideration, remembering that any form may be so 
turned as to bring any desired side into a horizontal 
position to be used as a base, or that an upper hori- 
zontal surface can be used as a base as well as a lower. 
The operations of triangulation undoubtedly re- 
quire more care for the sake of accuracy than those of 
any other method of pattern cutting, for the reason that 
there is no opportunity of stepping off a continuous 
stretchout, at once, upon any line, either straight or 
curved. It is therefore not to be recommended if the j 



subject in hand admits of treatment by any regular 
method without too much subdivision. Triangulation 
is not introduced as an alternate method, but as a last 
resort, when nothing else will do. 

Besides the various forms of transition pieces, an- 
other class of forms is to be treated under this head, 
which might almost be considered as regular tapering 
articles. They include shapes, or frustums cut from 
shapes, which terminate in an apex, but whose bases 
cannot be inscribed in a circle, as irregular polygons, 
figures composed of irregular curves as well as the per- 
fect ellipse. A solid whose base is a perfect ellipse 
and whose apex is located directly over the center of 
its base (in other words, an elliptical cone) is perhaps 
the best typical representative of this class of figures. 
If the base of such a cone be divided into quarters by 
its major and minor axes, it will be seen at once that 
all of the points in the perimeter of any one quarter 
will be at different distances from the apex of the cone, 
because they are at different distances from the center 
of base or the intersection of the two axes. This is 
clearly shown in Fig. 266, in which are shown the two 
elevations and the plan of an elliptical cone. The side 
elevation shows K E to be the distance of the apex 
from the point P in the jslan of the base, while the end 
elevation shows K' D to be the distance of the apex 
from the point D of the base, or the true distance rep- 
resented by X D of the plan. 

If one-quarter of the plan of the base, as D P, be 
divided into any convenient number of equal spaces 
and lines be drawn to the center X, as shown, each line 
will represent the horizontal distance of a point in the 
perimeter from the apex ; and if a section of the cone 
be constructed upon any one of these lines, as, for in- 
stance, line 4 X, or, in other words, if a right angle 
triangle be drawn, of which 4 X is the base and Pi K 
the altitude, the hypothenuse will be the true distance 
of the point 4 from the apex. Therefore, to ascertain 
the distances from the apex to the various points in the 
circumference of the base construct a simple diagram 
of triangles, as shown in Fig. 267, viz. : Erect any 
perpendicular line, as X M, equal in hight to E K 
of the elevation; from X, on a horizontal line X P, as 
a base, set off the various distances of the plan, X 1, 
X 2, X 3, etc., numbering each point, and from each 
point draw a line to M. These hypothenuses will then 
represent the distances of the various points in the 
perimeter of the base from the apex of the cone ; or in 
other words, the sides of a number of triangles forming 
the envelope of the cone, the bases of which triangles 



92 



The New Metal Worker Pattern Book. 



will be the spaces 1 2, 2 3, etc., upon the plan. As all 
of these triangles terminate at a common apex or 
center, instead of laying out each one separately to 
form a pattern, as in the case of an article of the type 
shown in Fig. 261, the simplest method is as follows : 




pi^SiOlii 



Fig. 267.— Diagram of Sections on the Radial Lines of the Plan in 
Fig. 266, to which is Added the Pattern of One-quarter of the 

Envelope. 

From M, of Fig. 267, as a center, with radii corre- 
sponding to the distances from M to points on P X, as 
M 1, M 2, M 3, etc., describe arcs indefinitely, as 
shown to the left ; then taking the space used in step- 
ping off the plan between the points of the dividers, 
place one foot upon the arc drawn from point S, as at 
D, and swing the other foot around till it cuts the arc 
drawn from point 7 ; from this intersection as a center 
swing it around again, cutting the arc from 6 ; or in 
other words, step from one arc to the next till one- 
quarter of the circumference has been completed. 

As the spaces in the base are equal, it is clearly a 
matter of convenience whether this last ojDeration is 




Fig. 268. — Frustum of an Elliptical Cone. 

begun upon arc 8, stepping first to arc 7, then to arc 
6, etc., or whether it is begun upon arc 1, stepping 
first to arc 2, then to 3, etc., till complete. A line 
traced through these points, as A D, will give the cut 



at the base of the envelope, and ADM will be the 
envelope of one-quarter of the cone. 

In Fig. 26S is shown a perspective view of the 
frustum of the cone shown in Fig. 266, the upper sur- 
face A B being shown in Fig. 266 by the lines G H 
and X 0. If the envelope of such a frustum is desired 
the cut which its upper surface would make through 
the envelope of the entire cone could be obtained in 
exactly the same manner as that of its lower base, be- 
cause the upper surface of the frustum is in reality the 
base of the cone, which remains above after the lower 
part has been cut away. But as part of the operation 
has already been performed in obtaining the cut at the 
base, it is most easily accomplished as follows : First 
draw radial lines from the point M of the diagram of 




15:-=_-_-=-_^4T 



7 * \ \ \\\ 

* i i \"\Y\ 

5 1 \ \ \ V\ 



11 \ \\v> 

/// / / f L M J \°v \ \ A 

r u-h-4f 




Fig. 269.— Elevation of the Frustum of an Elliptical Cone. 

triangles, Fig. 267, to each of the points previously 
obtained in the cut at the bottom of the envelope, 
between A and D ; also draw a horizontal line at a 
hight above the base X P equal to E L, Fig. 266, cut- 
ting the hypothenuses M 1, M 2, etc., as shown by 
Gr BL Now place one foot of the dividers at the point 
M, and bringing the other foot successively to the 
various points of intersection of the line G H with the 
various hypothenuses, describe arcs cutting the radial 
lines in the envelope of corresponding number. A line 
traced through the points of intersection, as B C, will 
give the cut at the top of the envelope of the frustum, 
of which AD is the bottom cut. 

If the cut at the top of the frustum is to be oblique 
instead of horizontal, a means must be devised for 



Principles of Pattern Cutting. 



93 



measuring the distance from the apex at which the 
oblique plane cuts each of the hjpothenuses, or in other 
words, each of the lines drawn from the apex of the 
cone to the various points in its base. In Fig. 269, 
E S T F is the elevation of an oblique frustum of an 
elliptical cone, whose apex is at K, and whose base is 
the same and has been divided in the same manner as 
that shown in Fig. 266. 

Erect lines from each of the points in the curve of 
one-half the plan D A to the base line E F of the 
elevation, thenc^ carry them toward the apex K, cut- 
ting the line S T ; the vertical hight of the points upon 
S T can then most easily be measured by carrying them 
horizontally, cutting the center line R K of the cone, 
where to avoid confusion they should be numbered to 




Fig. 270. — Diagram of Sections on the Radial Lines of the Plan in 
Fig. 269, with the Pattern of One-half the Envelope. 

correspond with the points of the plan from which each 
was derived. These points may now be transferred in 
a body by any convenient means to the vertical line 
X/ M' of the diagram of triangles, Fig. 270, seeing that 
each point is placed at the same distance from M' that 
it is from the point K of Fig. 269. A horizontal line 
from any one of the points on the line X' M' extended 
to the hypothenuse of corresponding number will then 
give the correct distance of that point from the apex 
cf the cone. The diagram M' X' D' is a duplicate of 
M X P of Fig. 267, and the lower outline of the en- 
velope is the same as that shown in Fig. 267. It will 
be noted, however, that half the stretchout of the base 
is necessary in this case to give all the essentials of the 
pattern of the envelope, while one-quarter was sufficient 
for the previous operations. When all the points in 
the uppei line of the frustum have been obtained in the 



diagram they may be transferred to the various radial 
lines in the envelope, from M' as a center, by the use 
of the compasses as before, all as shown in the draw- 
ing. 

If the apex of the cone were not located directly 
over the crossing of the two axes of the ellipse — that is, 
if the cone were scalene or oblique instead of right — 
the method of obtaining its envelope, or parts of the 
same, would not differ from the foregoing. Lines 
drawn from the points of division in the circumference 
of the base to the point representing the position of 
the apex in the plan will be the horizontal distances 
used in constructing a diagram of triangles, which dis- 
tances can be used in connection with the vertical 
hight of the cone, as before, in obtaining the various 
hypothenuses. If the apex of a scalene cone be 
located over the line of either axis of the ellipse, either 
within the perimeter of the base or upon one of those 
lines continued outside the base, one-half the pattern 
of the entire envelope will have to be obtained at one 
operation ; but if the apex is not located upon either 
of those lines in the plan, then the entire envelope 
must be obtained at one operation, as no two quarters 
or halves of the cone will be exactly alike. 

The method of obtaining the envelope of any 
scalene cone, even though its base be a perfect circle, 
is governed by the same principles as those employed 
in the above demonstrations. 

It will be well to remember that any horizontal 
section of a scalene cone is the same shape as its base, 
which fact can be used to advantage in determining the 
best method to be employed in obtaining the envelope 
of any irregular flaring surface that may be presented. 
If, for instance, the plan of any article, whose upper 
and lower surfaces are horizontal, shows each to con- 
sist of two circles or parts of circles of different diame- 
ters not concentric, it is evident that the portion of the 
envelope indicated by the circles of the plan is part of 
the envelope of a scalene cone. An illustration of this 
is given in Fig. 271, which shows a portion of an 
article having rounded corners and flaring sides and 
ends, but with more flare at the end than at the side. 
The plan shows the curve of the bottom corner A B to 
be a quarter circle with its center at X, and that of the 
top CD to be a quarter circle with its center at Y. 
The rounded corner A B D G is then a portion of 
the envelope of a frustrum of a scalene cone, and the 
method of finding the dimensions of the complete cone 
is quite simple and is as follows : First draw a line, 
Z X, through the centers of the two circles in the plan, 



94 



Tlxe New Metal Worker Pattern Booh. 



at right angles to which project an oblique elevation, as 
shown below, making the distance between the two 
lines E F and G H equal to the hight of the article. 
Lines from X and M of the plan of the bottom fall 
upon G H, locating the points X' and H, while lines 
from Y and N of the top locate the points Y' and F in 
the upper line of the oblique elevation. A line drawn 
through Y' and X', the centers of the circles, will then 
represent the axis of the cone in elevation, which can 
be continued to meet a line drawn through the points 
F and II, representing the side of the cone, thus locat- 
ing the apex Z' of the scalene cone. The point Z' can 
then be carried back to the plan, as shown at Z, thus 
locating the apex in that view. As the line N Z rep- 
resents the horizontal distance between the point F and 
the apex Z' of the cone, so lines drawn from Z to any 
number of points assumed in the curve of the base 
C D will give the horizontal distances between those 
points and the apex, to be used as the bases in a dia- 
gram of triangles similar to that shown in Fig. 267, 
while V Z' gives their hight. Having drawn a dia- 
gram of triangles the pattern follows in the manner 
there shown. 

For greater accuracy in the case of a very tapering 
cone, the circles of the plan can be completed, as 
shown dotted, and their points of intersections with the 
line Z 1ST can be dropped into oblique elevation, as 
seen at S and T, through which a line can be drawn to 
meet a line through F and H with greater accuracy 
than one through Y' and X', as the angle in the former 
case is twice as great. 

In the above methods of obtaining the envelopes 
of what, may be termed irregular conical forms, it will 
be clearly seen that the operation of dividing the curve 
of the base into a great number of spaces really re- 
solves the conical figure into a many sided pyramid, 
and that the lines connecting the apex with the points 
in the base, which have been referred to as hypothe- 
nuses, are really the angles or hips of the pyramid. 
It is therefore self evident that any method of de- 
velopment which is applicable to a many sided pyra- 
mid is equally applicable to one whose sides are fewer 
in number, with the only difference, however, that the 
lines representing the angles or hips in the case of a 
pyramidal figure mean angles or sharp bends in the 
pattern of the envelope, while in the case of the conical 
envelope the bends are so slight as to mean only a con- 
tinuous form or curve. 

It is believed that the foregoing elucidation of the 
principles governing the development of the surfaces 



of irregular shaped figures is sufficiently clear to make 
the demonstrations of this class of problems, given in 
Chap. YI, Section 3, easily understood by the student, 
as well as to enable him to apply them to any new 
forms that may present themselves for solution. 

This chapter is intended to present, under its three 
different heads, all the principles necessary to guide 
the student in the solution of any problem that may 




Fig. 271.— Elevations and Plan of an Article, the Corner of which 
is a Portion of a Scalene Cone. 



arise. Its aim is to teach principles rather than rules, 
and the student is to be cautioned against arbitrary 
rules and methods for which he cannot clearly under- 
stand the reason. His good sense must govern him 
in the employment of principles and in the choice 
of methods. There is hardly a pattern to be cut which 
cannot be obtained in more than one way. Under 
some conditions one method is best, and under other 
conditions another, and careful thought before the 



Principles of Pattern Cutting. 



95 



drawing is begun will show which is best for the pur- 
pose in hand. 

The list of problems and demonstrations in the 
chapter which follows is believed to be so comprehen- 



sive that therein will be found a parallel to almost any- 
thing that may be required of the pattern cutter, and 
it is believed that he will have no difficulty in applying 
them to his wants. 



*m 



H 



Up 



CHAPTER VI. 



Pattern Problems 



Every effort has been put forth in the preceding 
chapters of this book to prepare the student for the all 
important work which is to follow — viz. , the solution 
of pattern problems. It is always advisable in the 
study of any subject to be well grounded in its funda- 
mental principles. For this reason a chapter on Linear 
Drawing has been prepared to meet the requirements of 
the student in pattern cutting, which is preceded by a 
description of drawing materials and followed by a 
solution of the geometrical problems of most frequent 
occurrence in bis work. But the most important 
chapter is the one immediately preceding this, in which , 
the theory of pattern cutting is explained, and which, 
if thoroughly understood, will render easy the solution 
of any problem the student may chance to meet. 

The selection of problems here presented is made 
sufficiently large and varied in character to anticipate, 
so far as possible, the entire wants of the pattern cut- 
ter, and the problems are so arranged as to be con- 
venient for reference by those who make use of this 



part of the book without previous study of the other 
chapters. 

In the demonstrations, only the scientific phase of 
the subject will be considered; consequently, all al- 
lowances for seams, joints, etc., as well as determining 
where joints shall be made, are at the discretion of 
the workman. In some of the problems it has been 
necessary to assume a place for a joint, but if the joint 
is required at a place other than where shown, the 
method of procedure would be slightly varied while 
the principle involved would remain the same. 

Each demonstration will be complete in itself, al- 
though references to other problems, principles, etc., 
will be made where such references will be of advan- 
tage to the student. 

As stated in the preceding chapter, the problems 
will be classed under three different heads according 
to the forms which they embody — viz. : First, 
Parallel Forms ; Second, Regular Tapering; Forms, and 
Third, Irregular Forms. 



SECTION 1 

Parallel ForoiSc 



(MITER CUTTING). 



The problems given in this section are such as 
occur in joining moldings, pipes and all regular con- 
tinuous forms at any angle and against any other form 
or surface, and in fact include everything that may 
legitimately be termed Miter Cutting. 

In the problems of this class two conditions exist, 
which depend upon the nature of the work. Accord- 
ing to the first, a simple elevation or plan of the inter- 
secting parts shows the miter line in connection with 
the profile, which is all that is necessary to begin at 
once with the work of laying out the patterns. 

It frequently happens, however, that moldings are 
brought obliquely against sloping or curved surfaces 



in such a manner that no view can be drawn in which 
the miter line will appear as a simple straight line. 
Hence it becomes necessary to produce by the inter- 
section of lines a correct elevation of the intersections 
of the various members of the molding, which when 
done results in the much sought miter line. Or it may 
be necessary to develop a correct profile of some oblique 
member or molding in order to effect a perfect miter. 
Thus some preliminary drawing must be done before 
the work of laying out the miter patterns can be prop- 
erly begun, which constitutes the second condition above 
referred to and forms the great reason why the pattern 
draftsman should understand the principles of projec- 



Pattern Problems. 



97 



tion, which have been simplified for his benefit in 
Chapter III. 

In the arrangement of the problems those which 
fulfill the first condition will precede those of the 
second, and all of a similar nature will, so far as pos- 
sible, be placed near together, so that the reader, 



knowing the kind that is wanted, will be able to find 
it with little difficulty. It will also be to his advantage 
before reading any of the problems in this chapter to 
read carefully the Eequirements and the general Rule 
governing this class of problems given in Chapter V 
on pages 76 and 77. 



PROBLEM I. 

A Butt Miter Against a Plain Surface Oblique in Elevation. 



Let A B L K in Fig. 272 be the elevation of a 
portion of a cornice, of which C D is the profile and 
A B the angle or inclination of the surface against 
which the cornice is required to miter. Divide the 
curved parts of the profile into spaces in the usual 
manner, and from all points in profile draw lines parallel 
with A K, cutting the miter line A B. On any con- 
venient line, as E F, at right angles to the cornice, lay 
off a stretchout of the profile C D, space by space as 
they occur, through the points in which draw the 
measuring lines, all as indicated by the small figures. 
Placing the T-square at right angles to the lines of the 
cornice, or, what is the same, parallel to the stretchout 
line, bring it successively against the points in the 
miter line A B and cut measuring lines of corresponding 
number, as indicated by the dotted lines. A line traced 
through these points, as indicated by H G, will be 
the pattern required. 




Fig. 272. — A Butt Miter Against a Plain Surface Oblique in Elevation 





i 13 

In 



Fig. 273 be the 



Fig. 27S.—A Butt Miter Against a Plain Surface Oblique in Plan. 



PROBLEM 2. 

A Butt Miter Against a Plain Surface Oblique in Plan. 

Let A B L K in 
plan of the cornice which is required 
to miter against a vertical surface stand- 
ing at any angle with the lines of the 
cornice, the angle being shown by A B. 
Draw the profile C D in position corresponding to the- 
lines of the cornice, all as indicated. Sj)ace the profile in 
the usual manner, and through the points draw lines 
jDarallel to the direction of the cornice, cutting the miter 
line A B. On any convenient line at right angles to the 
lines of the cornice lay off the stretchout E F of the jjro- 
file C D, through the points in which draw measuring 
lines in the usual manner. Placing the T-square at right 
angles to the cornice, or, what is the same, parallel to 
the stretchout line E F, bring it successively against the 
points in A B and cut the corresponding measuring lines. 
A line traced through the points of intersection thus ob- 
tained, shown by H G, will be the pattern required. 



98 The New Metal Worker Pattern Book. 

PROBLEM 3. 

A Square Return Miter, or a Miter at Right Angles, as in a Cornice at the Corner of a Building. 



In Fig. 274, let A B D C be the elevation of a 
cornice at the corner of the building for which a miter 
at right angles is desired. As has been explained in the 
chapter on the Principles of Pattern Cutting (page 77), 
the process of cutting a miter for a right angle admits 
of certain abbreviations not employed when other 




Fig. 274- — A. Square Return Miter, as in a Cornice at the 
Corner of a Building. 

angles are required. The demonstration here intro- 
duced is calculated to show the method of obtaining 
the pattern for a square miter with the least possible 
labor. Divide the profile A B into any convenient 
number of parts, as shown by the small figures. At 
right angles to the lines of the molding, and in con- 
venient proximity to it, lay off the stretchout E F, 



through the points in which draw measuring lines in 
the usual manner, parallel to the lines of the cornice, 
producing them far enough to intercept lines dropped 
vertically from points in A B. Place the T-square at 
right angles to the cornice, or, what is the same, parallel 
to the stretchout line, and, bringing it successively 
against all the points in the profile A B, cut measuring 
lines of corresponding numbers. Then a line traced 
through these points, as shown by G H, will be the 
pattern sought. The reason for this is as follows : As 
the angle of this miter cannot be shown in any other 
view than a plan, the plan is the correct view from 
which to derive the pattern ; having drawn which, as 



■7 



ZJ 



Fig. £75. — Plan of a Square Return Miter. 

shown in Fig. 275, the operation of developing the 
pattern becomes exactly the same as in the previous 
problem (Fig. 273). In Fig. 274, A B D C represents 
the elevation of a portion of a cornice, while A B rep- 
resents the profile of the return or receding portion 
against which the piece A B D C is required to miter, 
or, in other words, the miter line. As the profiles of 
the face piece and of the return piece are of course the 
same, the outline A B becomes at once the profile and 
the miter line ; therefore that portion of the rule which 
says, " drop the points from the profile on to the miter 
line," must be omitted. All that remains then is to 
drop the points at once into the stretchout. 



PROBLEM 4. 

A Return Miter at Other Than a Right Angle, as in a Cornice at the Corner of a Building. 



In Fig. 276, let A B C D be the elevation of a 
portion of cornice, and let G H K be the plan of any 
angle around which the cornice is to be carried, a pattern 
being required for an arm of the miter. Complete 



the plan by drawing the lines E F and F L, inter- 
secting at F, giving the correct projection of the mold- 
ing from G H and H K, and then draw the miter line 
between the joints H and F. It will be observed that 



Pattern Problems. 



99 



the arm G H F E has been projected directly from the 

profile A B, thus 
placing profile and 
plan in correct rela- 
tion to each other. 
Divide the profile A 
B in the usual man- 
ner into any conven- 




-A Return Miter at Other than a Right Angle, as 
in a Cornice at the Corner of a Building. 



ient number of parts, and from the points thus ob- 
tained drop lines vertically on to the miter line in the 
plan F H, as shown. At right angles to this arm of 
the cornice, as shown in plan, lay off a stretchout of 
the profile, as shown by N" M, through the points in 
which draw the usual measuring lines, as indicated. 
Place the T-square parallel to this line, or, what is 
the same, at right angles to E F, and, bringing it suc- 
cessively against the points in F H, cut 
measuring lines of corresponding num- 
bers. Then a line traced through the 
points thus obtained, as shown by O P, 
will be the pattern sought. As inti- 
mated at the outset of this problem, the 
angle GHK represents any angle what- 
ever, and the course to be pursued is 
exactly the same whether it be acute 
or obtuse. Of course the more acute 
the angle GHK the longer will the 
miter line H F become, as may be ascer- 
tained by experiment, producing a cor- 
responding increase in the projection of 
the different parts of the pattern from 
the line N M. 



PROBLEM 5. 



A Butt Miter Against a Curved Surface. 



In Fig. 277, let A B be the pi-ofile of any cornice 
and D K H C be the elevation of the same showing the 
curved surface D, against which it is required to 
miter. The principle herein involved is exactly the 
same as that in Problem 1. Space the profile in the usual 
manner, and through the points draw lines cutting C D. 
At right angles to the line of cornice lay off the stretch- 
out L M, as shown, through the points in which 
draw measuring lines in the usual manner. Place the 
J-square parallel to the stretchout line, or, what is the 
same, at right angles to the lines of the cornice, and, 
bringing it against the several points in C D, cut the 
corresponding measuring lines, as shown. In the 
event of a wide space, as shown by a' b 1 in the eleva- 
tion, the curve between these points may be trans- 
ferred to the pattern by means of a piece of tracing 
paper, or, if it is a regular curve, its radius may be 
used as shown by G and G' and the arrow points. A line 
traced through the several points of intersection, as 
shown by E F, will be the shape of the required pattern. 




Fig. 277.— A Butt Miter against a Regular Curved Surfaae. 



100 



The New Metal Worker Pattern Book. 




278. — The Patterns for a Hip Finish in a Curved Mansard Roof , the Angle of the Hip being a Right Angle. 



Pattern Problems. 



101 



PROBLEM 6. 



The Pattern for a Hip Finish in a Curved Mansard Roof, the Plan of the Hip Being a Right Angle. 



The solution of all problems concerning mansard 
roofs, and especially those in which the roof surface is 
curved, calls for much good judgment on the part of 
the pattern cutter, for the reason that the original 
designs that come into his hands are seldom drawn 
mathematically correct. The upper part of a mansard 
dome, such as is shown in Fig. 278, as it curves 
away from the eye, becomes so much flattened in 
appearance that, if drawn correctly, it might, to any 
but an expert draftsman, create a false impression of 
the design intended ; hence the original drawing must 
often be taken for what it means rather than for what 
it says. 

The engraving represents an elevation of a curved 
hip molding occurring in a roof, of which E D is the 
vertical hight and M 2 K 3 is a section. The first step 
to be described is the method of obtaining the pattern 
of the fascias of the hip molding. For this purpose is 
shown in the drawing such a representation of it as 
would appear if the two fascias formed a close joint 
upon the angle of the roof, supposing that the hip 
molding or the bead is to be added afterward on the 
outside over this joint. The part to be dealt with may 
be considered the same as though it were the section 
of a molding, instead of a section of a roof, and the 
operations performed are identical with those employed 
in cutting a square miter. Space the profile H K into 
any convenient number of parts, introducing lines in 
the upper part in connection with the ornamental 
corner piece, shown b} r L D, at such intervals as will 
make it possible to take measurements required to 
describe the shape of it in the pattern. From this 
profile, by means of the points just indicated, lay off a 
stretchout, as shown by H 1 K 1 , and through the points 
draw the usual measuring lines. Bring the T-square 
against the several points in II Iv, and cut the corre- 
sponding lines drawn through the stretchout just 
described. Then a line traced through these points, 
as shown by H 2 K 2 , will be the outside line of the 
fascia. For the inside line take the given width of the 
fascia and set it off from this line at intervals, measur- 
ing at right angles to it, as indicated by A 1 B 1 , and 
not along the measuring lines of the stretchout, as 
would be indicated by A 1 C. Then a line traced 
through these points, as shown from M 1 to L 1 , will be 



the inside line of the fascia strip. The points in the 
ornamental corner piece from L 1 to D 1 are to be 
obtained from the elevation, in case a correct elevation 
is furnished the pattern cutter, by measurement along 
the lines drawn horizontally through the several points 
in L D, which are transferred to the measuring 
lines of corresponding number in the stretchout already 
referred to. Or the shape from L 1 to D' may be 
described arbitrarily upon the pattern at this stage of 
the operation, according to the finish required upon 
the roof. The latter method is the preferable one. The 
method of constructing the elevation, by working back 
from the outline thus established, is clearly indicated 
by the dotted lines in the engraving. From the 
several points in the profile II K horizontal lines are 
drawn, as shown, and from the intersections of the 
inside line of the pattern of the fascia piece with the 
various measuring lines, as above described, lines are 
dropped, cutting these horizontal lines of correspond- 
ing numbers. Then a line traced through these points, 
shown from M to L, will be the inside line of the 
fascia piece in elevation. To cut the flange strip 
bounding the fascia and corner piece, commonly called 
the sink strip, an elevation of which is shown in the 
section from M 2 to D 2 , the following method will be 
the simplest, and at the same time sufficiently accurate 
for all purposes : Draw the line Gr F approximately 
parallel to the upper part of the section M 2 ~D% making 
it indefinite in length, which cut by lines drawn from 
the several points in M 2 D 2 , at right angles to it, as 
shown. From F G, upon the several lines drawn at 
right angles to it, set off spaces equal to the distance 
upon lines of corresponding number from D E to the 
line M L of the elevation. Then a line traced through 
these points, as indicated by M 3 1/, will constitute a pro- 
file of this flange strip. In like manner set off in contin- 
uation of it, the lengths measured from points in the 
ornamental corner piece to D E, all as shown by L 3 D 3 F. 
From this profile lay off a stretchout parallel to G F, as 
shown by M 4 D 4 , through the points in which draw- 
measuring lines in the usual manner. Place the 
T-square parallel to the stretchout line, and, bringing 
it successively against points in both the inner and the 
outer lines of the elevation of the flange strip, as 
shown from M 2 D 2 , cut the measuring lines of corre- 



102 



The New Metal Worker Pattern Book. 



sponding number. Then lines traced through these 
points of intersection, as shown from M" to D", will be 



the pattern of the flange strip bounding the edge of 
the fascia. 



PROBLEM 7. 



Miter Between Two Moldings of Different Profiles. 



To construct a square miter between moldings of 
dissimilar profiles requires two distinct operations. 
The miter upon each piece is to be cut as it would ap- 
pear when intersected by the other molding. Let A B 
and A 1 B 1 in Figs. 279 and 280 be the profiles of two 
moldings, between which a square miter is required. 
As, of course, the two arms of the miter are different, 
it will be necessary to draw an elevation of each show- 
proper outline against which it is to miter, 
with the profile A B, project 
from it an elevation, as shown by FCDE, Fig. 279, 
terminating such elevation by the profile of the other 
molding, A 1 B", as shown by F E. Then, as in the 
case of Problems 1 and 6, the line F E becomes the 



ing the 

Beginning, therefore, 



through these points, as shown by F l E 1 , will give the 
shape qf the cut on the piece A B to fit against the 
profile E F. For the other piece proceed in the same 
manner, reversing the order of the profiles. From its 
profile A 1 B 1 produce the elevation K M N L, Fig. 280, 
completing same by means of the profile of the first 
molding, A B, as shown by M N. Divide A 1 B' m 
the usual manner. Through the points draw lines cut- 
ting M N. At right angles to this piece lay off the 
stretchout P of the profile A 1 B 1 , through the points 
in which draw measuring lines, as shown. With the 
J-square at right angles to the line K M or N L, and 
brought against the points in M 1ST, cut corresponding 
measuring lines drawn through OP. A line traced 




Fig. £79. 




Fig. 280. 



Miter between Two Moldings of Different Profiles. 



miter line, and the method of procedure is the same 
as in those problems. Divide A B into any convenient 
number of parts in the usual manner, from which carry 
lines horizontally against F E. At right angles to the 
lines of the molding lay off a stretchout, G H, of the 
profile A B, through the points in which draw the 
usual measuring lines. Bring the T-square against 
the points of intersection in the line E F, and cut the 
corresponding measuring lines. Then a line traced 



through these points, as shown by M 1 W, will be the 
shape of the end of the piece required to fit against the 
profile M N. In the event of the points obtained by 
spacing the profiles A B and A 1 B 1 not meeting all the 
points in the profiles F E and M N necessary to be 
marked in the pattern, then lines must be drawn back- 
ward from such points in profiles M N" and E F, cut- 
ting the profile A 1 B 1 or A B, as the case may be. 
Corresponding points are then to be inserted in the 



Pattern Problems. 



103 



stretchouts, through which measuring lines are to be 
drawn, which, in turn, are to be intersected by lines 
dropped from the points. An illustration of this occurs in 
Fig. 280, where it will be seen that no point obtained by 
the dividing of the profile A' B 1 strikes the point X of 
the miter line, which is absolutely necessary to the 
shape of the pattern. Therefore, after spacing the 
profile, a line is drawn from X back to A 1 B 1 , foring the 
point No. 6^-. In turn this point is transferred to the 



stretchout P, also marked 6£, from which a measur- 
ing line is drawn in the same manner as through the 
other points in the stretchout, upon which a point from 
X is dropped, as shown by X 1 . In actual practice such 
expedients as this must be resorted to in almost every 
case, because usually there is less correspondence be- 
tween the members of dissimilar profiles, between which 
a miter is required, than in the illustration here given. 
By this means profiles, however unlike, can be joined. 



PROBLEM 8. 



A Butt Miter Against an Irregular or Molded Surface. 



Let B A in Fig. 281 be the profile of a cornice, 
against which a molding of the profile, shown by G II, is 
to miter, the latter meeting it at an inclination, as indi- 




Fig. 281. — A Butt Miter against an Irregular or Molded Surface. 

cated by C D. Construct an elevation of the oblique 
molding, as shown by C D F E, in line with which 
draw the profile G H. Divide G H in the usual man- 



ner into any convenient number of parts, and through 
the points draw lines parallel to the lines of the in- 
clined molding, cutting the profile B A, all as indicated 
by the dotted lines. At right angles to the lines of the 
molding, of which a pattern is sought, lay off a stretch- 
out, M N, in the usual manner, through the points in 
which draw measuring lines. Place the T-square at 
right angles to the lines of the inclined molding, or, 
what is the same, parallel to the stretchout line, and, 
bringing it against the points of intersection formed by 
the lines drawn from the profile G H across the profile 
B A, cut the corresponding measuring lines. In the 
event of any angles or points occurring in the profile 
B A which are not met by lines drawn from the points 
in G H, additional lines from these points must be 
drawn, cutting the profile G H, in order to establish 
corresponding points in the stretchout. Thus the 
points 3 and 13 in the profile G H are inserted after 
spacing the profile, as described in Problem 7, because 
the points with which they correspond in the profile 
B E are angles which must be clearly indicated in the 
pattern to be cut. Having thus cut the measuring 
lines corresponding to the points in the profile B A,- 
draw a line through the points of intersection, as shown, 
by P. Then P will be the shape of the pattern of 
the incline cornice to miter against the profile A B. 



104 



The New Metal Worker Pattern Booh. 

PROBLEM 9. 

The Pattern of a Rectangular Flaring: Article. 



In Fig. 282, let C A B E be the side elevation of 
the article, of which F I K M is the plan at the base 
and G II L N" the plan at the top. Let it be required 
to produce the pattern in one piece, the top included. 
Make II 1 L 1 N l G 1 in all respects equal to II L N" G of 
the plan. Through the center of it likewise draw RP 
indefinitely, and through the center in the opposite 
direction draw S indefinitely. From the lines IP L 1 
and G 1 N 1 set off T O and W S respectively, each in 
length equal to the slant hight of the article, as shown 
by G A or E B of the elevation. Through and S 
respectively draw I 1 K 1 and F 1 M 1 , parallel to II 1 L 1 and 
G 1 N 1 , and in length equal to the corresponding sides in 
the plan I K and F M, placing one-half that length 
each way from the points and S. In like manner set 
off V P and TJ R, also equal to C A, and draw through 
R and P the lines F 2 I 2 and K 2 M 2 , parallel to the ends 
of the pattern of the top part as already drawn, and in 
length equal to I F and K M of the plan. Draw I 1 H 1 , 
K' L 1 , K 2 IV, W N 1 , M' N 1 , F 1 G', F 2 G' and P H 1 , 
thus completing the pattern sought. In the same 
general way the pattern may be described, including 
the bottom instead of the top, if it be required that 
■way. 

Considering this problem in the light of miter cut- 
ting proper, I H G F and F G N M may be regarded as 
the plan of two similar moldings of which A C is the 
profile, I II, G F and N M being the miter lines. T 
is the stretchout line, drawn at right angles to F 
M, while I 1 K' and H' L' are the measuring lines 
representing respectively the points C and A of the 
profile. 



The points F and G are then dropped into their 
respective measuring lines, thus locating the points I 1 



ELEVATION. 




Fig. 2S2. — The Pattern of a Rectangular Flaring Article. 

and IF at one end of the pattern, while points K 1 and 
L 1 are derived from M and N at the other end. 



PROBLEM 10. 



Patterns of the Face and Side of a Plain Tapering: Keystone. 



Let A B D C in Fig. 283 be the elevation of the 
face of a keystone, and G E 2 F 2 K of Fig. 284 a sec- 
tion of the same on its center line. 

Sometimes problems occur which are so simple 
that it is not apparent that their solution is an ex- 
emplification of any rule. That this, with others in 
which plain surfaces form the largest factors, may be 



so designated, will be sufficient excuse for a brief ref- 
erence to first principles. This problem is generally- 
referred to as finding the "true face" of the keystone, 
because, the face being inclined, the elevation ABC 
D does not represent the ' ' true face " or " true ' ' 
dimensions of the face. To state the case, then, in con- 
formity with the rule, A B and C D are the upper and 



Pattern Problems. 



105 



lower lines of a molding, of which E J F 5 of Fig. 284: is 
the profile, and A C and B D are the surfaces against 




Fig. 283. 
Patterns of the Face and Side of a Plain Tapering Keystone. 

which it miters, or the miter lines. Therefore, to lay 
out the pattern, draw any line, as E' F', at right angles 
to A B for a stretchout line, upon which lay off the 
stretchout taken from the profile W F 3 , Fig. 284, which 



in this case consists of only one space, as shown by 
E 1 F' ; through the points E 1 and F 1 draw the hori- 
zontal lines A 1 B' and C D 1 , which are none other than 
the measuring lines. Then, with the T-square placed 
parallel with the stretchout line, drop the points from 
the miter lines A C and B D into lines of correspond- 
ing letter, which connect, as shown by A 1 C 1 and B 1 D 1 , 
which completes the pattern. 

In developing the pattern for the side, W G and 
F 3 K are the lines of the molding, B D of Fig. 283 its 
profile and E 3 F 3 the miter line. Hence upon any 
vertical line, as L K 1 , lay off the stretchout of profile 
B D, locating the points M' and IT, all as shown by 
L M H l Iy 1 , through which points draw the measuring 
lines ; then, with the T-square placed parallel to L K', 
drop the points E 3 and F 3 into lines of corresponding 
letter, as shown by E 3 F\ As the vertical lines at Gl- 
and K represent the position of surfaces against which 
the side is required to fit at the back, bring the T-square 
against each, thus locating them in the pattern at G' 
and K', as shown. 

As the side must also fit over the molding of the arch 
an opening must be cut in it corresponding in shape 
to the profile of the arch molding 1ST, which is given in 
the sectional view. It is therefore only necessary to 
transfer this profile to the pattern, placing the top at 
the measuring line M and the bottom at the measuring 
line H', all as shown at N 1 . 



PROBLEM II. 

Patterns for the Corner Piece of a Mansard Roof, Embodying: the Principles Upon Which All Mansard 

Finishes are Developed. 



One of the first steps in developing the patterns 
for trimming the angles of a mansard roof is to obtain 
a representation of the true face of the roof. In other 
words, inasmuch as the surface of the roof has a slant 
equal to that shown in the profile of the return, the 
length of the hip is other than is shown in the eleva- 
tion, and this difference in dimensions extends in a pro- 
portionate degree to the lines of the various parts form- 
ing the finish. Not only are the vertical and oblique 
dimensions different, but, as the result of this, the 
angle at A is different from that shown in a normal ele- 
vation. Hence, it is of the greatest importance to ob- 
tain a " true face " or elevation of the roof as it would 



appear if swung into a vertical position, which may be 
accomplished as follows : 

In Fig. 285, let A E F C be the elevation of a 
mansard roof as ordinarily drawn, and let A 1 G be the 
profile showing the pitch drawn in line with the eleva- 
tion. Set the dividers to th< length A 1 Gf, and from 
A 1 as center, strike the arc G Gr 1 , letting G 1 fall in a 
vertical line from A 1 . From G 1 draw a line parallel to 
the face of the elevation, as shown by G 1 G 1 , and from 
the several points in the hip finish, as shown by 
C and K, drop lines vertically, cutting G" C in the 
points C 1 and K 1 , as shown. From these points carry 
lines to corresponding points in the upper line of the 



106 



The New Metal Worker Pattern Book. 



elevation, as shown by C 1 A and K 1 h. Then A C F 1 

E represents the pattern of the surface shown by A 
F E of the elevation. In cases where the whole hight 
of the roof cannot be put into the drawing for use, as 



B 1 draw the horizontal line, as shown by B' B', and 
from B drop a vertical line cutting this line, as shown, 
in the point B\ By inspection of the engraving it will 
be seen that the point B 3 falls in the line A C obtained 




Elevation ' 

Fig. 285.— The Plain Surfaces of a Mansard Roof Developed. 



above described, the same result may be accomplished 
by assuming any point as far from A as the size of the 
drawing will permit, as B, and treating the part between 
A and B as though it were the whole. That is, from 
A, in a vertical line, set off A B", equal to A B. From 



in the previous operation, thus demonstrating that the 
latter method of obtaining the angle by which to pro- 
portion the several parts results the same as the method 
first described, and therefore may be used when more 
convenient. 



PROBLEM 12. 

A Face Miter at Right Angles, as in the Molding Around a Panel. 



In Fig. 286, let A B D C represent any panel, 
around which a molding is to be carried of the profile 
at E and E 1 . The miters required in this case are of 
the nature commonly called "face" miters, to dis- 
tinguish them from other square miters, which can only 



be shown in a plan view. A correct elevation of the 
panel A B D C, with the lines of the molding carried 
around the same, determines the miter lines A F and 
G C, which, in connection with the profiles at E and E 1 
are all that is necessary to the development of the pat- 



Pattern Problems. 



107 



tern. The two profiles are here drawn, thus constitut- 
ing an entire section of the panel, because it is usual, 
for constructive reasons, to cut the two moldings with 
the intervening panel in one piece where the width of 




Fig. 286. — A Face Miter at Right Angles, as in the Molding Around 

a Panel. 



the metal will permit it. Divide the two profiles in 
the usual manner into the same number of parts, from 
which points draw lines parallel to the lines of the 
molding, cutting the miter lines, as shown. For the 



pattern of the side corresponding to A B lay off a 
stretchout at right angles to it, as shown by H K, 
through which draw measuring lines in the usual man- 
ner. Place the T-square at right angles to A B, or, 
what is the same, parallel to the stretchout line Ii K, 
and, bringing it successively against the several points 
in the miter line A F, cut measuring lines of corre- 
sponding number. Then a line traced through these 
points, as shown by L M, will be the pattern sought. 
The other pattern is developed in like manner. It is 
usual to draw the stretchout lines, K H and K 1 H' 
across the lines of the moldings which they represent, 
beginning the stretchouts at the inner lines of the mold- 
ing, thus : Point 10 of profile E would be located at V, 
while point 10 of profile E 1 would be at ~W. While this 
is apt to produce some confusion of lines in actual 
practice, it gives the entire profile in one continuous 
stretchout for the purpose alluded to above — that of 
cutting the entire width of the panel in one piece. 
Should it be desired to make one of the moldings 
separate from the rest, an additional point for the pur- 
pose of a lap is assumed at one of the moldings, as 11 
of profile E 1 . The pattern for the end piece, A C, 
may be derived without drawing an additional profile, 
as its profile and stretchout are necessarily the same as 
that of the other two arms ; therefore reproduce H K 
on a line at right angles to A C, as shown by JST O, 
through the points in which draw measuring lines in 
the usual manner, producing them sufficiently far in 
each direction to intercept lines dropped from the 
points in the two miter lines. Place the T-square at 
right angles to A C, and, bringing it successively 
against points already in A F and C G, cut measuring 
lines of corresponding numbers. Then lines traced 
through the intersections thus formed, as shown by 
P R and S T, will be the shape of the pattern of the 
end piece. 

It may be noticed in the last operation that drop- 
ping the points from either of the miter lines, as A F, 
into the measuring lines is, in fact, only continuing in 
the same direction the lines previously drawn from the 
profile E to the line A F; and that in reality the shape 
of the cut at P R is developed without the assistance 
of the miter line, thus giving another instance of the 
fact that any square miter can be cut by the short 
method when the relation of the parts is understood. 



108 



The New Metal Worker Pattern Booh. 

PROBLEM 13, 

The Patterns of the Moldings Bounding; a Panel Triangular in Shape. 



In Fig. 287, let D E F be the elevation of a 
triangular panel or other article, surrounding which is a 
molding of the profile, shown at Gr and Gr 1 . Construct 
an elevation of the panel molds, as shown by ABC, 
and draw the miter lines A D, B E and C F. For the 
patterns of the several sides proceed as follows : Draw 
a profile, Gr, placing it in correct relative position to 
the side D F, as shown. Divide it into any con- 



of the three sides, at convenient points, draw stretch- 
out lines, as shown by H I, H 1 I 1 and H 2 F, through 
the points in which draw the usual measuring lines. 
With the T-square parallel to each of the several 
stretchout lines, or, what is the same, at right angles 
to the respective sides, bringing the blade successively 
against the points in the several miter lines, cut the 
corresponding measuring lines, all as indicated by the 




Fig. 287.— The Patterns of the Moldings Bounding a Triangular Panel. 



venient number of parts in the usual manner, and 
through these points draw lines, as shown, cutting the 
miter lines F C and A D. In like manner place the 
profile Gr 1 in a corresponding position relative to the 
side E F. Divide it into the same number of parts, 
and draw lines intersecting those drawn from the first 
profile in the line F C, also cutting the line E B. 
By this operation points are obtained in the three 
miter lines A D, E B, F C, from which to lay off the 
patterns in the usual manner. At right angles to each 



dotted lines. Then lines traced through the points of 
intersection thus obtained will describe the patterns 
required. A 1 C 1 F 1 D 1 will be the pattern for the side 
A D F C of the elevation, and likewise C 2 B 2 E 2 
F 2 is the pattern for the side described by similar 
letters. 

Placing another profile in the molding A B D E 
would, if divided the same as the others, only result 
in another set of intersections at the points already 
existing on the lines A D and B E, as occurred on 



Pattern Problems. 



109 



the line F C, hence to save labor one profile in this 
case is all that is really necessary, the points being 



carried around from that and dropped into the three 
stretchouts respectively. 



PROBLEM 14. 

The Patterns of a Molding: Mitering Around an Irregular Four-Sided Figure. 



In Fig. 288, let A B C D be the elevation of an ir- 
regular four-sided figure, to which a molding is to be 
fitted of the profile shown by K. Place a duplicate 
profile against the side opposite, as shown, from which 



in these several stretchouts draw measuring lines in the 
usual manner, producing them until they are equal in 
length to the respective sides, the pattern of which is 
to be cut. Placing the T-square at right angles to the 




Fig. 288.— The Patterns of a Molding Mitering Around an Irregular Four-Sided Figure. 



project the lines necessary to complete the elevation 
of the molding as it would appear when finished, all as 
shown by E F G H. Draw the several miter lines 
B F, C G, D H and A E. Divide the two profiles into 
the same number of parts in the usual manner, through 
the points in which draw lines parallel to the lines of 
the molding in which they occur, cutting the miter 
lines, as shown. At right angles to each of the several 
sides lay off a stretchout from the profile, as shown by 
L M, L 1 M 1 , L 3 M 2 , L 3 M 3 . Through the several points 



lines of the several sides, or, what is the same, parallel 
to the stretchout lines, bring it against the points in 
the miter lines, cutting the corresponding measuring 
lines, all as indicated by the dotted lines. Then the 
lines traced through these points of intersection will 
give the several patterns required. Thus E' H 1 D 1 A 1 
will be the pattern of the side E H D A of the eleva- 
tion ; B? D" C G 1 will be the pattern of the side 
H D C G ; G 3 F' B 1 C 3 that of F B C G ; and F 3 B 3 
A 3 E 3 that of the remaining side. 



110 



The New Metal Worker Pattern Book. 

PROBLEM 15. 



The Patterns of Simple Gable Miters. 



In Fig. 2S9, let A B K and BKEbe the angles 
of the miters at the foot and peak of a gable. Draw 
profiles of the required molding in correct relation to 
both the horizontal and inclined moldings, as shown 
at II and H 1 , through the angles of which draw 
the other parallel lines necessary to complete the ele- 
vation. Their intersection at the base of the gable 
produces the miter line B C, while the miter line at 
the top of the gable is a vertical line, because the two 
sides of the gable, K B and K K, are of the same 
pitch. The profile H is so placed as also to repre- 
sent the return at the side at its proper distance 
from B. Divide the profile H in the 
usual manner into any convenient num- 
ber of equal parts. Place the T-square 
parellel to the lines in the horizontal 
molding, and, bringing it successively 
against the points in the profile, cut the 
miter line B C, as shown. At right 
angles to the lines of the horizontal 
cornice draw the stretchout E F, through 
the points in which draw the usual 
measuring lines, as shown. Reverse 
the T-square, letting the blade lie parallel 
to the stretchout line E F, and, bringing 
it against the several points of the profile 
H, cut the corresponding measuring 
lines. Then a line traced through these 
points of intersection, as shown from G 
to V, will be the pattern of the end of 
the horizontal cornice mitering with 
the return. In like manner, with the 
T-square in the same position, bring it • 
against the points in the miter line B C, 
and cut corresponding measuring lines drawn through 
the same stretchout. Then a line traced through the 
points of intersection thus obtained, as shown by T IT, 
will be the pattern of the end of the horizontal cornice 
mitering against the inclined cornice. Divide the pro- 
file II 1 into any convenient number of equal parts, all as 
indicated by the small figures. Through these points 
draw lines cutting the miter line B C, and also the 
miter line K L at the top. At right angles to the 
lines of the raking cornice place a stretchout, E 1 F 1 , of 



the profile H 1 , through the points in which draw the 
usual measuring lines, as shown. Place the T-square 
parallel to this stretchout line, and, bringing it suc- 




Fig. 389.— The Patterns of Simple Gable Miters. 

cessively against the points in B C and K L, cut the 
corresponding measuring lines, all as indicated by the 
dotted lines. Through the points thus obtained trace 
lines, as indicated by M N and P. Then M N will 
be the pattern for the bottom of the raking cornice 
mitering against the horizontal, and P will be the 
pattern for the top of the same. The pattern shown 
at G V will also be the pattern for the return miter- 
ing with A D of the elevation, it being necessary only 
to reverse it and to establish its length. 



Pattern Problems. 

PROBLEM 16. 



ill 



The Pattern for a Pedestal of Which the Plan is an Equilateral Triangle. 




should be drawn so as to show one side in profile and 
the plan placed to correspond with it. Draw the miter 
lines E and G 0. Divide the profile B D into spaces 
of convenient size in the usual manner, and number 
them as shown in the diagram. From the points thus 
obtained drop lines, cutting E and G 0, as shown. 
Lay off the stretchout N P at right angles to the side 
E G, and through the points in it draw measuring lines. 
Place the J-square at right angles to E G, and, bring- 
ing it successively against the points in the miter lines 
E O and G 0, cut the corresponding measuring lines. 
A line traced through these points will be the pattern, 
as shown by H L M K. 

The principle involved in this and several follow- 
ing problems is exactly the same as that of the preced- 
ing regular and irregular shaped panels. In this case 
the shape of the article is shown in plan instead of ele- 



Pattern 



Fig. 290.— The Pattern for a Pedestal of which the Plan is an Equilateral Triangle. 



Let A B D C in Fig. 290 be the elevation of a 
pedestal or other article of which the plan is an equi- 
lateral triangle, as shown by F E G. This elevation 



vation, and the profile is too large to permit of its 
being drawn within the plan, as were the profiles of the 
panel moldings in their elevations. 



112 



The New Metal Worker Pattern Booh. 



PROBLEM 17. 

The Pattern for a Pedestal Square in Plan. 

In Fig. 291, let A B D C be the elevation of a 
pedestal the four sides of which are alike, being in 
plan as shown by E H G F, Fig. 292. Since the plan 
is a rectangular figure the miters involved are square 
miters, or miters forming a joint at 90 degrees. A 
square miter admits of certain abbreviations, the rea- 
sons for which are explained in Problem 3, as well as 
in Chapter V, under the head of Parallel Forms. The 
abbreviated method which is here illustrated is always 
The plan is introduced only to show the shape 



used. 



of the article, and is not employed directly in cutting 




Fig. 292.— The Plan of Square Pedestal. 

the pattern. Space the profiles, shown in the eleva- 
tion by A C and B D, in the usual manner, numbering 
the points as shown. Set off a stretchout line, L R, at 
right angles to the base line C D of the pedestal, 
through the points in which draw measuring lines. 
Place the T-square parallel to the stretchout line, and, 
bringing it successively against the points in the two 
profiles, cut the corresponding lines drawn through the 
stretchout. A line traced through these points, as 
shown by L M N K, will be the pattern of a side. 




Fig. 291.— The Pattern for a Pedestal, Square in Plan. 



PROBLEM 18. 

The Patterns for a Vase, the Plan of Which is a Pentagon. 



In Fig. 293, let S C K T be the elevation of a 
vase, the plan of which is a pentagon, as shown C 1 
C R P. The elevation must be drawn in such a man- 



ner that one of the sides will be shown in profile. 
Draw the plan in line and in correspondence with it. 
Divide the profile into spaces of convenient size in the 



Pattern Problems. 



113 



usual manner and number them. Draw the miter 
lines C 1 H 1 and C 2 H 2 in the plan, and, bringing the 
T-square successively against the points in the profile, 
drop lines across these miter lines, as shown by the 
dotted lines in the engraving. Lay off the stretchout 
M N at right angles to the piece in the plan which 



the case of a complicated profile, or one of many dif- 
ferent members, to drop all the points across one sec- 
tion of the plan C H' H 2 Q" would result in confusion. 
Therefore it is customary, in practice, to treat the 
pattern in sections, describing each of the several 
pieces of which it is composed independently of the 




Fig. 293.— Pattern for the Upper Part. 
The Patterns for a Vase, the Plan of which is a Pentagon. 



corresponds to the side shown in profile in the eleva- 
tion. Through the points in it draw the usual measur- 
ing lines. Place the T-square parallel to the stretch- 
out line, and, bringing it against the several points in 
the miter lines which were dropped from the elevation 
upon them, cut the corresponding measuring lines 
drawn through the stretchout. A line traced through 
the points thus obtained will describe the pattern. In 



others. In the illustration given the pattern has been 
divided at the point H, the upper portion being 
developed from the profile and plan, as above, while 
the lower part is redrawn in connection with a section 
of the plan, as shown in Fig. 294. Corresponding 
letters in each of the views represent the same parts, 
so that the reader will have no trouble in perceiving 
just what has been done. Instead of redrawing a por- 



114 



The New Metal Worker Pattern Booh. 



tion of the elevation and plan, as has been done in this 
case, sometimes it is considered best to work from one 
profile rather than to redraw a portion of it, as that 
always results in more or less inaccuracy. Therefore, 
after using the plan and describing a part of the 
pattern, as shown in the operation explained above, 



a piece of clean paper is pinned on the board, cover- 
ing this plan and pattern, upon which a duplicate 
plan is drawn, from which the second section of 
the pattern is obtained. Great care, however, is neces- 
sary in redrawing portions of the plan to insure 
accuracy. 



PROBLEM 19. 



The Pattern for a Pedestal, the Plan of Which is a Hexagon. 



4 




sides, drawn so that one of the sides will be shown in 
profile. Place the plan below it and corresponding 
with it. Divide the profile shown in the elevation 
into any convenient number of spaces in the usual 
manner, and, to facilitate reference to them, number 
them as shown. Bring the T-square against the points 
in the profile and drop lines across one section of the 
plan, as shown by H X M. At right angles to this 
section of the plan lay off the stretchout line N 0, 
through the points in which draw the usual meas- 
uring lines. Place the T-square parallel to the 
stretchout line, and, bringing it successively against 
the points in the miter lines H X and M X, cut the 
corresponding measuring lines, as indicated by the 
dotted lines. Then a line traced through the points 



Fig. 295.— The Pattern for a Pedestal, the Plan of which is a Hexagon. 



In Fig. 295, let C D F E be the elevation of a 
pedestal which it is desired to construct of six equal 



thus obtained will be the required pattern, as shown by 
P STE. 



Pattern Problems. 



115 



PROBLEM 20. 

The Pattern for a Vase, the Plan of Which is a Heptagon. 




sides will be shown in profile. In line with it draw the 
plan, placing it so that it shall correspond with the 
elevation. Space the profile L P in the usual manner, 
and from the points in it drop lines crossing one sec- 
tion of the plan, cutting the miter lines K S and II V, 
as shown. Lay off a stretchout, A B, at right angles 
to the side of the plan corresponding to the side of the 
vase shown in profile in the elevation. Through the 
points in it draw the usual measuring lines. Place the 
T-square parallel to this stretchout line, and, bringing 
it successively against the points in the miter lines, 
cut the corresponding measuring lines, as shown. A 
line traced through these points, as shown by K O 



Fig. 296. — The Pattern for a Vase, the Plan of which is a Heptagon. 

In Fig. 296, let E L P G be the elevation of the I "W U, will be the pattern of one of the sides of the 
vase, constructed in such a manner that one of its | vase. 



PROBLEM 21. 

The Patterns for an Octagonal Pedestal. 



Let K H G W L in Fig. 297 be the elevation of 
a pedestal octagon in plan, of which the pattern of a 
section is required. This elevation should be drawn 
in such a manner that one side of it will appear in 
profile. Place the plan so as to correspond in all 
respects with it. Divide the profile G W, from which 
the plan of the side desired is projected, in the usual 
manner, and from the points in it drop points upon 
each of the miter lines F T and P U in the plan. Lay 
off a stretchout, B E, at right angles to the side of the 
plan corresponding to the side of the article shown in 
profile in the elevation, and through the points in it 



draw the usual measuring lines. Place the T-square- 
parallel to the stretchout line, and, bringing it suc- 
cessively against the points dropped upon the miter 
lines from the elevation, cut the corresponding measur- 
ing lines. A line traced through the points thus 
obtained will describe the pattern of one of the sides 
of which the article is composed. In cases where the 
profile is complicated, consisting of many members, 
and where it is very long, confusion will arise it all 
the points are dropped across one section of the plan, 
as above described. It is also quite desirable in many 
cases to construct the pattern in several pieces. In 



118 



The New Metal Worker Pattern Book. 



such cases methods which are described in connection 
with Problem 18 may be used with advantage. 
In the present case the pattern is constructed of two 



the pattern is cut by means of a part of the plan 
redrawn above the elevation, thus allowing the use 
of the same profile for both. The same letters refer 



F_ 




Fig. 297.— The Patterns for an Octagonal Pedestal. 



pieces, being divided at the point S of the profile. 
The lower part of the pattern is cut from the plan 
drawn below the elevation, while the upper part of 



to similar parts, so that the reader will have no diffi • 
culty in tracing out the relationship between the dif- 
ferent views. 



Pattern Problems. 

PROBLEM 22. 

The Patterns for a Newel Post, the Plan of Which is a Decagon. 



117 



In Fig. 29S, let WTJSPOKTVbe the eleva- 
tion of a newel post which is required to be constructed 
in ten parts. Draw the plan below the elevation, as 
shown. The elevation must show one of the sections 





«s 



23 p 



the plan, as shown by G X II , and cutting the two 
miter lines G X and H X. Lay off the stretchout line 
C D at right angles to Gr II, and through it draw the 
customary measuring lines. Place the T-square parallel 
to the stretchout, and, bringing it against the several 
points in the miter lines G X and H X, cut the corre- 
sponding measuring lines. A line traced through the 
points thus obtained will describe the pattern. In order 
to avoid confusion of lines, which would result from drop- 
ping points from the entire profile across one section of 
the plan, a duplicate of the cap A 1 W is drawn in Fig. 
299 in connection with a section of the plan, as shown 




Fig. 299.— Pattern of Cap. 




*3D 



Fig. 298. — The Patterns for a Neivel Post, the Plan of which is a Decagon. 



or sides in profile, and the plan must be placed to cor- 
respond with the elevation. Space the molded parts of 
the profile in the usual manner, and from the points in 
them drop lines crossing the corresponding section of 



by Gr 1 X' H 1 , which are employed in precisely the same 
manner as above described, thus completing the pattern 
in two pieces, the joint being formed at the point num- 
bered 11 of the profile and the stretchout. 



118 



The New Metal Worker Pattern Book. 

PROBLEM 23. 

The Patterns for an Urn, the Plan of Which is a Dodecagon. 



In Fig. 300, let X A G II be the elevation of an 
urn to be constructed in twelve pieces. The elevation 
must be drawn so as to show one side in profile. Con- 
struct the plan, as shown, to correspond with it and 



the several points in the miter lines N X and O X, 
cut the corresponding measuring lines. A line 
traced through the points thus obtained will describe 
the pattern sought. In this illustration is shown a 
method sometimes resorted to by pattern cutters to 
avoid' the confusion resulting from dropping all the 
points across one section of the plan. The points from 
13 to 20 inclusive are dropped upon the line X. 
The stretchout C D is drawn in exactly the middle of 
the pattern — that is, it is drawn from X, the central 
point of the plan. Points are transferred by the J- 
square from X to the measuring lines on one side of 




V Plan yj 

Fig. SOO.—The Patterns for an Urn, the Plan of which is a Dodecagon. 



draw the miter lines. Divide the profile A S G into 
spaces in the usual manner, and from the points thus 
obtained drop lines across one section, N X 0, of the 
plan. Lay off the stretchout C D at right angles to the 
side N of the plan. Place the T-square parallel to 
the stretchout, and, bringing it successively against 



the stretchout, the points on the other side being ob- 
tained by duplicating distances from C D on the several 
lines. The points 1 to 13 are dropped on N X only. 
The stretchout E F is laid off at right angles to the 
side M N from the point X, and, the T-square being set 
parallel to E F, the points are transferred to the 



Pattern Problems. 



119 



measuring lines on one side of E F, while the distances 
on the opposite side are set off by measurement, as de- 



scribed in the first instance. This plan will be found ad- 
vantageous in complicated and very extended profiles. 



PROBLEM 24. 

The Pattern for a Drop Upon the Face of a Bracket. 



In Figs. 301 and 302, methods of obtaining the 
return strip fitting around a drop and mitering against 
the face of a bracket are shown. Similar letters in 
the two figures represent similar parts, and the follow- 



II K, as shown by P, and on P lay off a stretch- 
out, through the points in which draw the usual 
measuring lines. From the points in the profile F Gr 
carry lines in the direction of the molding — that is, 





The Pattern for a Drop upon the Face of a Bracket. 



ing demonstration may be considered as applying to 
both. Let A B D C be the elevation of a part of the 
face of the bracket, and H K L a portion of the side, 
showing the connection between the side strip of the 
drop E F Gr and the face of the bracket. To state the 
case simply, F Gr is the profile and N" M the miter 
line, because N M is the outline of the surface against 
which the side strip miters. Then, following the rule, 
divide F Gr into any convenient number of parts in the 
usual manner, as shown by the small figures. Produce 



parallel to K M — intersecting the face of the bracket 
N M. Eeverse the T-square, placing the blade parallel 
to the stretchout line O P, and, bringing it suc- 
cessively against the points in 1ST M, cut the cor- 
responding measuring lines, as indicated by the 
dotted lines. Then a line traced through these several 
points of intersection, as shown by E' P, will 
be the pattern of the strip fitting around E F Gr 
and mitering against the irregular surface N M of 
the bracket face. 



120 



The Isew Metal Worker Pattern Book. 



PROBLEM 25. 
The Pattern of a Boss Fitting; Over a Miter in a Molding*. 



Let A B in Fig. 303 be the part elevation of a 
pediment, as in a cornice or window cap, over the 
miter and against the molding and fascia in which a 
boss, FKGH, is required to be fitted, all as shown 
by N E D of the side view. 

The outline F K G H of the boss is to be con- 
sidered as the profile of a molding running in the direc- 
tion shown by D E in the side view, and mitering 
against the surface of the cornice shown by IS!" E. 
For the 'natterns proceed as follows: Divide so much 
of the profile of the boss K F H G as comes against 
the cornice, shown from K to F, into any convenient 
number of parts, and from these points draw lines 
parallel to D E — that is, to the direction of the mold- 
ing under consideration — until they intersect the 
miter line N E, which in this case is the profile of 
the cornice molding. As the boss is so placed over 
the angle in the cornice molding that the distance from 
K to F is the same as that from K to G, the part of 
the boss K G will be an exact duplicate of K F and 
may be duplicated from the pattern of K F without 
another side view drawn especially for it, which would 
have to be done if the boss was otherwise placed. 
Therefore, extend the line N D upon which to lay off 
a stretchout of Iv F H G, dividing the portion K 1 F 1 
into the spaces shown at K F of the profile, through 
which draw the usual measuring lines. Make the jDor- 
tion F 1 G' equal in length to the part F H G and, 
lastly, the portion G' K a a duplicate of F' K 1 reversed, 
as shown. Place the T-square parallel to the stretch- 



out line K 1 K", and, bringing it against the several 
points in N 0, cut corresponding measuring lines, as 




Fig. SOS. — The Pattern of a Boss Fitting Over a Miter in a Molding. 

shown. Then lines traced through these points of in- 
tersection, as shown by K 1 L M K 3 , will be the re- 
quired pattern. 



PROBLEM 26. 

The Patterns for a Keystone Having- a Molded Face With Sink. 



In Fig. 304, let E A B F be the front elevation 
of a keystone, as for a window cap, of which KLMP 
S R is a sectional view, giving the profile of the mold- 
ing M N P, over which it is required to fit. The 
sink in the face extends throughout its entire length, 
and is shown by G H D C, its depth being shown by 
the line K T of the section. E F H G and A B D C 
thus become moldings, of which E A and F B are the 
parallel lines, E F, G II, C D and A B the miter lines, 
and K R the profile. Likewise C G H D becomes 
a molding, of which G BZ and C D are the miter 
lines and K T the profile. Therefore, to obtain the 
pattern of the face pieces, divide the profile of the 
face K R into any convenient number of spaces, 



and from the points thus obtained carry lines across 
the face of the keystone, as shown. At right angles to 
the top of the keystone lay off a stretchout of K R, as 
shown by K 3 R 1 , through which draw the usual 
measuring lines. Placing the T-square parallel to the 
stretchout line, and bringing it successively against 
the points in the lines C D and A B bounding the face 
strip, cut the corresponding measuring lines. Then a 
line traced through these points, as shown by C 3 A 3 B 3 
D 3 , will be the pattern for this part. 

In developing the pattern for the sink the usual 
method would be to divide K T into equal spaces, 
carrying lines across the face, and thence into the 
stretchout : but since this would result in confusion of 



Pattern Problems. 



121 



lines, the same points as were established in K E have 
been used, which are quite as convenient as the others 
mentioned, save that the points in K T must be ob- 
tained from the points in K B, by carrying lines back 
to K T, as shown, and in laying off the stretchout each 
individual space must be measured by the dividers. 

K 2 c 3 




S04. — The Patterns for a Keystone Having a Molded Face 
with Sink. 

At right angles to the line H D of the keystone lay off 
a stretchout of K T, as shown by K 1 T 1 , through the 
points in which draw the usual measuring lines. Place 
the T-square at right angles to the lines across the face 
of the keystone, and, bringing it successively against 
the points in the lines G H and C D, forming the sides 
of the sink, cut the corresponding measuring lines 
drawn through K 1 T 1 . Then lines traced through these 
points, as indicated by Gr 1 H l and C 1 D', will form the 



pattern of the required sink piece. For the pattern 
of the piece forming the sides of the sink in the 
face of the keystone, K E T becomes the elevation 
of a molding running in the direction of E T, of which 
K E and Iv T are the miter lines and C D the profile. 
Hence, at any convenient place above or below the 
sectional view, lay off the stretchout of the line C D, 
as determined by the lines drawn across it in the first 
operation, all as indicated by C 2 D\ Through the 
points in C D" draw measuring lines in the usual man- 
ner. The next operation, in course, would be to drop 
lines from the points in the profile to the miter lines ; 
but as this has already been done by the lines of the 
first operation, it is only necessary to place the J- 
square at light angles to the measuring lines, and 
bring it successively against the several points in the 
lines K E and K T, and cut the corresponding measur- 
ing lines, as shown. Then a line traced through these 
points, as indicated by K 3 B 2 and K 3 T 2 , will be the 
pattern of the piece required. 

For the side of the keystone, KLSR becomes 
the face of a molding, of which A B is the profile and 
K E the miter line at one side, and L M and P S the 
miter lines at the other. From this point forward the 
problem is, in principle, the same as Problem 10. 
For convenience, and to avoid confusion, it is best to 
asain make use of the same set of lines instituted in the 
first part of the demonstration. Therefore, lay off the 
stretchout A 1 B l equal to A B, putting into it all the 
points occurring in A B, through which draw measur- 
ing lines in the usual manner. Place the J-square at 
right angles to these measuring lines, and, bringing it 
Successively against the points in the line Iv E, and 
likewise against L M and P S of the back, cut corre- 
sponding measuring lines, as shown. Then a line 
traced through these points of intersection, as shown 
by W M 1 L 1 K 4 E 2 S 1 P 1 O 1 , will be the outline of the 
required pattern, with the exception of that part lying 
between 1ST and O 1 , which make a duplicate of N" 0. 
By examination of the points in A 1 B 1 and the lines 
drawn through the same and making comparison with 
the points in A B, it will be seen that in order to locate 
accurately the position of the profile of the window cap 
molding MSOP, two additional points, as shown by a; 1 
and?/ 1 , have been introduced, corresponding to x and y, 
the points of intersection between the extreme lines of 
the cap molding itself and the side of the keystone 
A B, as shown in the elevation by the curved lines of 
that molding. In practice it is frequently necessary to 
introduce extra points in operations of this character. 



122 



The New Metal Worker Pattern Book. 

PROBLEM 27. 

The Pattern of a Square Shaft to Fit Against a Sphere. 



In Fig. 305, let H A A' K be the elevation of a 
square shaft, one end of which is required to fit against 
the ball D F E. Draw the center line FL, upon which 
locate the center of the ball G. Continue the sides of 
the shaft across the line of the circumference of the 
ball indefinitely. From the points of intersection be- 
tween the sides of the shaft and the circumference of 
the ball, A or A', draw a line at right angles to the 
sides of the shaft, across the ball, cutting the center 
line, as shown at B. Set the dividers to G B as radius, 
and from G as center, describe the arc C C, cutting 
the sides in the points C and C. Then HCC'K will 
be the pattern of one side of a square shaft to fit 
against the given ball. 




Fig. S05. The Pattern of a Square Shaft to Fit Against a Sphere 



PROBLEM 28. 



To Describe the Pattern of an Octagon Shaft to Fit Against a Ball. 



Let H F K in Fig. 306 be the given ball, of which 
G is the center. Let D" C C 3 D 3 E represent a plan of 
the octagon shaft which is required to fit against the 
ball. Draw this plan in line with the center of the 
ball, as indicated by F E. From the angles of the 
plan project lines upward, cutting the circle and con- 
stituting the elevation of the shaft. From the point A 
or A 1 , where the side in profile cuts the circle, draw a 
line at right angles to the center line of the ball F E, 
cutting it in the point B, as shown. Through B, from 
the center G by which the circle of the ball was 
struck, describe an arc, cutting the two lines drawn 
from the inner angles C 2 C 3 of the plan, as shown at 
C and C 1 . Then MCC'N will be the pattern of one 
side of an octagon shaft mitering against the given 
ball H F K. If it be desired to complete the eleva- 
tion of the shaft meeting the ball, it may be done by 
carrying lines from C and C horizontally until they 
meet the outer line of the shaft in the points D and 
D 1 . Connect C 1 and D 1 , also C and D, by a curved 
line, the lowest point in which shall touch the hori- 
zontal line drawn through B. Then the broken line 
D C C D 1 will be the miter line in elevation formed 
by the junction of the octagonal shaft with the ball. 




Fig. S06.—The Pattern of an Octagon Shaft to Fit Against a Ball. 



Pattern Problems. 



123 



PROBLEM 29. 

The Patterns of an Octagonal Shaft, the Profile of Which is Curved, Fitting over the Ridge of a Root. 



In Fig. 307 is shown the elevation and plan of the 
shaft of a finial of the design shown in Fig. 30S. The 
shaft is octagon throughout, and if it were designed to 
stand upon a level surface, the method of obtaining its 
patterns would be the same in all respects as that de- 




F Gr E of the plan, into any number of parts in the 
usual manner, and from these points carry lines verti- 
cally crossing the miter lines Gr E and G F. From the 
center Gr draw E' M' at right angles to E F, upon which 
line lay off a stretchout of the profile J L, drawing 

measuring lines through the 
points. Place the T-square, 
parallel to the stretchout line, 
and, bringing it successively 
against the points in G E and 
G F, cut corresponding meas- 



uring lines, as shown, and 



Fig. 307.— Plan, Elevation and Patterns. 

The Patterns of an Octagonal Shaft, Curved in Profile, Fitting 
over a Ridge. 

scribed in Problem 21. As shown by the line KmK, 
however, its lower end is designed to fit over the ridge 
of a roof or gable, to obtain the patterns of which pro- 
ceed as follows : 

Construct a plan of the shaft at its largest sec- 
tion, as shown by A B C D E F, from the center of 
which draw miter lines, as shown by Gr E and G F. 
Divide the profile of the shaft J L, corresponding to 



through the points thus ob- 
tained trace lines, all as indi- 
cated in the drawing. This 
gives the general shape of the 
pattern for the sides of the shaft. By inspection of the 
plan and elevation together, it will be seen that to fit the 
shaft over the roof some of the sections composing it 
will require different cuts at their lower extremities. 
Two of the sections will be cut the same as the 
pattern already described. They correspond to the 
side marked A B and E F in the plan. Two others, 
indicated in the plan by CD and H I, will be cut to 
fit over the ridge of the roof, as shown in the elevation 
by n m n. The remaining four pieces, shown in plan 
by B C, D E, F I and A H, will be cut obliquely to fit 
against the pitch of the roof, as shown by n in the 
elevation. For the sides C D and H I, shown in the 
center of the elevation, it will be seen that the line 
drawn from 1 touches the ridge in the point m, while 
the line drawn from 3 corresponds to the point at which 
the side terminates against the pitch of the roof. 
Therefore, in the pattern draw a line from the center 
of it, on the measuring line 1, to the sides of it, on the- 
measuring line 3, all as shown by m 1 n 1 and in u'. 
Then these are the lines of cut in the pattern corre- 
sponding to m n and m n of the elevation. By further 
inspection of the elevation, it will be seen that for the 
remaining four sides it is necessary to make a cut in 
the pattern from one side, in a point corresponding to 
3 of the profile, to the other, in a point corresponding 
to 1 of the profile, all as shown by n 0. Taking corre- 
sponding points, therefore, in the measuring lines of 
the pattern, draw the lines n' o\ as shown. Then the 
original pattern, modified by cutting upon these lines, 
will constitute the pattern for the four octagon sides. 



124 



The New Metal Worker Pattern Booh. 



PROBLEM 30. 

To Construct a Ball in any Number of Pieces, of the Shape of Gores. 



Draw a circle of a size corresponding to the- re- 
quired ball, as shown in Fig. 309, which divide, by any 
of the usual methods employed in the construction of 
polygons, into the number of parts of which it is 
desired to construct the ball, in this case twelve, all as 
shown by E, F, G, H, etc. From the center draw 
radial lines, R E and R F, etc., representing the joints 
between the gores, or otherwise the miter lines. If the 
polygon is inscribed, as shown in the illustration, it will 
be observed that the joint or miter lines will lie in the 
surface of the sphere and that therefore the middle of 
the pieces, as shown at W, C and u\ will fall inside 
the surface of the sphere a greater or less distance 
according to the number of gores into which the 
sphere has been divided, and that therefore it becomes 
necessary to construct a section through the middle 
of one of the sides for use as a profile from which to 
obtain a stretchout. It will be well to distinguish 
here between absolute accuracy and something that 
will do practically just as well and save much labor. 
This profile, if made complete, would have for its 
width the distance W u\ while its hight or distance 
through from R to a point opposite would be equal to 
the diameter of the circle, or twice the distance R II. 
As one-quarter of this section will answer every pur- 
pose, it may be constructed with sufficient accuracy as 
follows : Supposing R E F to be the piece under con- 
sideration, draw a line parallel to its center line R C 
conveniently near, as A V, upon which locate the 
points A and V by projection from C and R, as 
shown by the dotted lines. From the point V erect 
the line B V perpendicular to V 1 A, and make B V 
equal to the radius of the circle, or R V ; then an arc 
of a circle cutting the points B and A will complete 
the section. This can be done by taking the radius 
R U between the points of the compasses and describ- 
ing an arc from the point V, whose distance from V 
is equal to the distance u l TJ. To develop the pat- 
tern divide B A into any convenient number of equal 
parts, and from the divisions thus obtained carry 
lines across the section E R F at right angles to a 
line drawn through its center, and cutting its miter 
lines, all as shown in R E and R F. Prolong the 
center line R C, as shown by S T, and on it lay off 
a stretchout obtained from B A, through the points 
in which draw measuring lines in the usual manner. 
Place the T-square parallel to the stretchout line, and, 



bringing it successively against {he points in the 
miter lines R E and R F, cut the corresponding 
measuring lines, as shown. A line traced through 
these points will give the pattern of a section. If, on 
laying out the plan of the ball, the polygon had been 
drawn about the circle, instead of inscribed, as shown 
in the engraving, it is quite evident that a quarter of 




Fig. S09. — To Construct a Ball in any Number of Pieces, of 
the Shape of Gores. 

the circle would have answered the purpose of a profile. 
These }3oints, with reference to the profile, are to be 
observed in determining the size of the ball. In the 
illustration presented, the ball produced will corre- 
spond in its miter lines to the diameter of the circle laid 
down, while if measured on lines drawn through the 
center of its sections it will be smaller than the circle. 

The patterns for a ball made up of zones or strips 
having parallel sides will be found in Section 2 of 
this chapter (Regular Tapering Forms). 



Pattern Problems. 

PROBLEM 31. 

The Pattern of a Round Pipe to Fit Against a Roof of One Inclination. 



125 




Fig. SlO—The Pattern of a Round Pipe to Fit Against a Roof of 
One Inclination. 



In Fig. 310, let A B be the pitch of the roof and 
C F D E the profile of the pipe which is to miter 
against it. Let Gr P H be the elevation of the pipe 
as required. Draw the profile in line with the eleva- 
tion, as shown by C F D E, and divide it into any con- 
venient number of equal parts. Lay off a stretchout 
in the usual manner, at right angles to and opposite 
the end of the pipe, as shown by I K, and draw the 
measuring lines. Place the T-square parallel to the 
sides of the pipe, and, bringing it successively against 
the divisions of the profile, cut the pitch line, as shown 
by A B. Eeverse the T-square, placing it at right 
angles to the pipe, and, bringing it successively against 
the points in A B, cut the corresponding measuring 
lines. A line traced through the points thus obtained, 
as shown by L M N, will finish the pattern. 



PROBLEM 32. 
The Pattern of an Elliptical Pipe to Fit Against a Roof of One Inclination. 



In Fig. 311, let NCDO be the elevation of an 
elliptical pipe fitting against a roof, represented by 
A B. Let E F G Q be the section or profile of the 
pipe. Draw the profile in convenient proximity to the 
elevation, as shown, and divide it into any convenient 
number of equal parts. Place the T-square parallel to 
the sides of the pipe, and, bringing it against the points 
in the profile, drop lines cutting the roof line A B, as 



through the points in it draw measuring lines in the 
usual manner. Eeverse the J-square, placing it at 




Fig. SU.—The Pattern of an Elliptical Pipe to Fit Against a Roof of One Inclination. 



shown. Opposite to the end of the pipe, and at right 
angles to it, lay off a stretchout, as shown by II I, and 



right angles to the pipe, and, bringing it successively 
against the points in A B, cut the corresponding 



126 



The New Metal Worker Pattern Booh. 



measuring lines, as indicated. A line traced through 
these points, as shown by K L M, will be the required 
pattern. In the illustration the long diameter of the 
ellipse, or E G, is shown as crossing the roof. The 
method of procedure would be exactly the same if the 
pipe were placed in the opposite position — that is, with 



the short diameter Q F crossing the roof. In such case 
the profile should be turned so that Q F is across the 
roof, or parallel to C D, and the elevation duly pro- 
jected from it. The pipe might with equal facility be 
placed so that the long diameter should lie at any 
oblique angle desired. 



PROBLEM 33. 



The Pattern of an Octagon Shaft Fitting: Over the Ridge of a Roof. 



In Fig. 312, let A B C be the section and D II 
G I E the elevation of an octagon shaft mitering 
against a roof, represented by the lines F G and G K. 
Place the section in line with the elevation, as shown, 
and from the angles drop lines, giving T V and U 
W of the elevation. Drop the point G back on to 
the section, thus locating the points 9 and i. Opposite 
the end of the shaft, and at right angles to it, draw a 
stretchout line, as shown by S R, and through the 
points in it draw measuring lines in the usual man- 
ner. Place the T-square at right angles to the shaft, 
and, bringing it successively against the points in the 



G, representing the ridge of the roof, cut the corre- 
sponding measuring lines. Then a line traced through 
the points thus obtained, all as shown by P N M L 





Fig. S12.—The Pattern of an Octagon Shaft Fitting Over the Ridge of a Roof. 



roof line formed by the intersection with it of the 
angle lines in the elevation, and also against the point 



in the engraving, will be the lower end of the pattern 
required. 



Pattern Problems. 



127 



PROBLEM 34 

The Pattern of a Round Pipe to Fit Over the Ridge of a Roof. 

Let A B C in Fisr. 313 be a section of the roof and 



D S B T E an elevation of the pipe. Draw a profile of 
the pipe in line, as shown by F 
G H. Since both inclinations 
of the roof are to the same angle, 




Fig. SIS. — The Pattern of a Round Pipe to Fit Over the Ridge of a Roof. 

both halves of the pattern will be the same. There- 
fore space off the half of the profile which miters 



against one slope of the roof, and lay off the stretch- 
out of the same upon the stretchout line I K, drawn 
at right angles to the lines of the pipe, which may 
be duplicated in a reverse order for the other half, 
as shown. Draw measuring lines through these points 
in the usual manner. Place the 
T-square parallel to the sides of 
the pipe, and, bringing it against 
the points in the profile, cut the roof 
line, as shown from B to T. Beverse 
the T-square, placing it at right 
angles to the lines of the pipe, and, 
bringing it successively against the 
points dropped upon the roof line, 
cut the corresponding measuring 
lines. A line traced through the 
points, as shown by L M N 
P, will form that end of the pattern which meets 
the roof. 



PROBLEM 35. 

An Octagon Shaft Mitering Upon the Ridge and Hips of a Roof. 

In Fig. 314 are shown the front and side elevations 
of a hipped roof, below which are placed plans, each 
turned so as to correspond with the elevation above it. 



Before the pattern of the shaft can be developed it will 
be necessary to obtain a correct elevation of its inter- 
section with the roof. Therefore, considering the plan 




Fig. S14.—The Pattern of an Octagon Shaft Fitting Over the Ridge and Hips of a Roof. 



An octagonal shaft is required to be mitered down 
upon this roof, so that its center line or axis shall inter- 
sect the apex of the roof C, as shown upon the plans. 



of the shaft as the profile of a molding, number all its 
points in both plans, beginning, for convenience, at the 
ridge of the roof, and including the points where the 



128 



The New Metal Worker Pattern Book. 



oblique sides cross the hips of the roof, as shown by 
the small figures 1 to 11. The next step is to project 
lines upward into the elevations from each of these 
points, continuing them till they intersect the lines of 
the roof, as shown by the vertical dotted lines. From 
each of these intersections in either view lines can be 
projected horizontally to the other view till they in- 
tersect with lines of corresponding number. Thus the 
points 9 and 10 cut the line of the hip in the front ele- 
vation at the point B, which, being carried across to 
the side view and intersected with lines from points 9 
and 10 from the plan below it, give the correct posi- 
tion of those points in the side view. In like manner 
the intersection of lines from points 6 and 7 in the side 
view, with the hip line at D, give the correct hight of 
those points iu the front view. Points 5 and S, being 
upon the hips, must appear in the elevations at points 
where the vertical lines from them cut the hip lines in 



the elevations. Lines connecting these points (5, 6, 7 
and 8, and 8, 9, 10 and 11) will complete the eleva- 
tions. In case all sides of the roof have the same pitch 
and the shaft is a regular octagon, all the angles of the 
shaft except 2 and 11 will intersect the roof at the 
same hight, in which case it will only be necessary to 
draw the front view. But should the slope of the front 
of the roof be different from that of the sides, it will be 
necessary to follow the course above described. To 
develop the pattern, draw any horizontal line, as E F, 
upon which place the stretchout of the octagon shaft 
obtained from the plan, as shown by the small figures, 
through which draw the usual measuring lines at right 



angles to it, and intersect the 



measuring lines 



with 



lines of corresponding numbers drawn horizontally 
from the intersections in the elevation. A line traced 
through these intersections, as shown by X Y Z, will 
be the desired pattern. 



PROBLEM 36. 

The Pattern of a Flange to Fit Around a Pipe and Over the Ridge of a Roof. 



In Fig. 315, let A B G be the section of the roof 
against which the flange is to fit, and let P S B be 
the elevation of the pipe required to pass through the 
flange. Let the flange in size be required to extend 
from A to G over the ridge B. Since both sides of the 
roof are of the same pitch, both halves of the opening 
from the point B will be the same. Therefore, for 
convenience in obtaining both halves of the pattern at 
one operation, the line B C may be continued across 
the pipe toward A 1 , and used in place of B A, the dis- 
tance from B on either line to the side B being the 
same. Under these conditions it will be seen that the 
process of describing the pattern is identical with that 
in the previous problem. Make B A 1 equal to B A, 
and proceed in the manner described in the problem 
just referred to. Divide the profile DEFG into any 
number of equal parts in the usual manner, and from 
the points so obtained carry lines vertically to the line 
A' C, and thence, at right angles to it, indefinitely. 
Also carry lines in a similar manner from the points A 1 
and C. Draw II L parallel to A 1 C. Make H I the 
width of the required flange, and draw I K parallel to 
H L. Through that part of the flange in which the 
center of the required opening is desired to be draw 
the line A 5 C, crossing the lines drawn from the pro- 
file. From each side of this line, on the several 




Fig. S15. — The Pattern of a Flange to Fit Around a Pipe and Over 
the Ridge of a Roof. 



Pattern Problems. 



129 



measuring lines, set off the same distance as shown 
upon the corresponding lines between D F and D 
E F, as shown. A line traced through the points 
thus obtained, as shown by D 1 E 1 F 1 G' 5 will be 



the required opening to fit the pipe. Through the 
center, across the flange, draw the line N M, which 
represents the line of bend corresponding to the ridge 
B of the section of the roof. 



PROBLEM 37. 



The Pattern of a Flange to Fit Around a Pipe and Against a Roof of One Inclination. 




Let L M, Fig. 316, be the inclination of 
the roof and P E T S an elevation of the 
pipe passing through it. N then represents 
the length of the opening which is to be cut 
in the flange, the width of which will be the 
same as the diameter of the pipe. Let A B 
D C be the size of the flange desired, as it 
would appear if viewed in plan. Immediately 
in line with the pipe draw the profile G H I 
K, putting it in the center of the plan of the 
flange A B D C, or otherwise, as required. 
Divide one-half of the profile in the usual 
manner, and carry lines vertically to the line 
L M, representing the pitch of the roof, and 
thence, at right angles to it, indefinitely. Carry 
points in the same manner from A and B. 
Draw C D 1 parallel to L M. Make C A' equal 
to A C, or the width of the required flange, 
and draw A 1 B' parallel to C 1 D\ Then C A« 
B 1 D 1 will be the pattern of the required flange. 
Draw E 1 F 1 through it at a point corresponding 
to E F of the plan, crossing the lines drawn 
from the profile. From E 1 F' set off on each 
side, on each of the measuring lines crossing it, • 
the width of opening, as measured on corre- 
sponding lines of the plan, measuring from E 
F in the plan to the profile. Through the 
points thus obtained draw a line, which will 
give the shape of the opening to be cut, all as 
shown by G' H' V K\ 



Fig. 315.— The Pattern of a Flange to Fit Around a Pipe and 
Against a Roof of One Inclination. 



130 



The New Metal Worker Pattern Book. 



PROBLEM 38. 



The Pattern for a Two-Piece Elbow. 



In Fig. 317, let A C B D be the profile of the pipe 
in which the elbow is to be made. Draw an elevation 
of the elbow with the two arms at right angles to each 
other, one of which is projected directly from the pro- 
file, as shown by E G I H K F. Draw the diagonal 
line G K, which represents the joint to be made. 
Divide the profile into any convenient number of equal 
parts. Place the T-square parallel to the lines of the 
arm of the elbow, opposite the end of which the profile 
has been drawn, and, bringing the blade successively 
against the several points in the profile, drop corre- 
sponding points on the miter or joint line K G, as shown 
by the dotted lines. Opposite the end of the same 
arm, and at right angles to it, lay off a stretchout line, 
M N, divided in the usual manner, and through the 
divisions draw measuring lines, as shown. Place the 
blade of the T-square at right angles to the same arm 
of the elbow, or, what is the same, parallel to the 
stretchout line, and, bringing it successively against 
the points in K G, cut the corresponding measuring 
lines, as shown. A line traced through these points, 
as indicated by E P 0, together with M N, will form 
the required pattern. 




Fig. S17.—The Pattern for a Two-Piece Elbow. 



PROBLEM 39. 



The Patterns for a Two-Piece Elbow in an Elliptical Pipe.— Two Cases. 



The only difference to be observed in cutting the 
patterns for elbows in elliptical pipes, as compared with 
the same operations in connection with round pipes, 
lies with the profile or section. The section is to be 
placed in the same position as shown in the rules for 
cutting elbows in round pipe, but it is to be turned 
broad or narrow side to the view, as the requirements 
of the case may be. In Figs. 318 and 319 are shown 
elevations and profiles of two right angled two-piece 
elbows in elliptical pipes. In Fig. 318 the broad side 
of the ellipse is presented to view, while Fig. 319 
shows the narrow side, as indicated by the respective 



positions of the profiles. Although the results in the 
two cases are different in consequence of the position 
of the profiles, the method of procedure is exactly the 
same. Similar parts in the two drawings have been 
given the same reference letters and figures, so that the 
following demonstrations will apply equally well to 
either : Let A C E F D B be the elevation of the 
elbow and H G K I its section. Draw C D, the 
miter line. Divide the profile ia the usual manner, 
as indicated by the small figures, and by means 
of the T-square placed parallel to the arm, drop 
points upon the miter line, as shown. Opposite the 



Pattern Problems. 



131 



end of the arm lay off a stretchout, M N, and through 
the points in it draw the usual measuring lines. 



points in the miter line, cut the corresponding meas- 
uring lines. A line traced through these points, 




Fig MA 




Fig. 819. 



.4 Two-Piece Elbow in Elliptical Pipe. 



Reverse the T-square, placing it at right angles to 
the arm, and, bringing it in contact with the several 



as shown by L P 0, will constitute the required 

miter. 



PROBLEM 40. 



The Patterns for a Three-Piece Elbow. 



In Fig. 320, let E M L I H K N F be the ele- 
vation of a three-piece elbow. The drawing of a 
three-piece elbow, at any angle whatever, should be 
so constructed that the middle section or portion bears 
the same angle with reference to the two arms. 
Since the two arms in the present instance are at right 
angles (90 degrees) to each other, the middle section 
must therefore be drawn at an angle of 45 degrees to 



both. Make its diameter the same as that of the two 
arms, and draw the miter lines M N" and L K. Draw 
the profile A B C in line with the arm from which the 
pattern is to be taken, as shown, and divide it into any 
convenient number of equal parts. Place the blade 
of the T-square parallel to this arm of the elbow, and, 
bringing it against the points in the profile, drop cor- 
responding points upon the miter line L K. At right 



1.32 



Tlie New Metal Worker Pattern Book. 



angles to L I draw a stretchout, as E S, through the 
divisions in which draw measuring lines in the usual 
manner. Placing the T-square at right angles to L I, 
and bringing it successively against the points 'in the 
miter line L K, cut the corresponding measuring lines, 



drop like divisions upon M N". At right angles to L 
M lay off a stretchout of the profile A B C, as shown 
by P 0, through the points in which draw measuring 
lines in the usual manner. Reversing the position of 
the set-square so that its long side shall come at right 





Fig. 320.— A Three-Piece Elbow. 



as shown. Then the line TUT, traced through the 
points thus obtained, forms in connection with S R 
the pattern of an end section. 

Place the 45-degree set square against the blade 
of the T-square so that its oblique or long side shall 
coincide with the lines of the middle section of the 
elbow, and, bringing it against the points in L K, 



angles to M L, or, what is the same, parallel to the 
stretchout line, bringing it successively against the 
several points in the miter lines M 1ST and L K, and 
cut the corresponding measuring lines. Then lines 
traced through these points, as shown by D X 
Y and G W Z, will be the pattern of the middle 
section. 



PROBLEM 41. 

The Patterns for a Four-Piece Elbow. 



In constructing the elevation of a four-piece 
elbow, first draw the profile ABC, from which pro- 
ject one of the arms of the elbow, as shown by the lines 
A F and C G, Fig. 321. At right angles to this lay 



off the other arm of the elbow, MLNI, continuing 
the lines of each until they intersect. Through the 
points of intersection draw the diagonal line a d. 
Establish the point a on this diagonal line at con- 



Pattern Problems. 



133 



venience, and from it draw the lines a b and a c at 
right angles to the two arms of the elbow respectively. 
From a as center, and with a b as radius, describe the 
arc bfe c, as shown, which divide into three equal 
parts, thus obtaining the points / and e. Through / 




Fig. SSI. — A Four-Piece Elboiv. 

and e, to the center a, draw the lines / a and e a, 
which will represent the centers of the middle sec- 
tions of the elbow, at right angles to which the sides 
of the same are to be drawn. Through/, and at right 



angles to fa, draw L K, meeting M L in the point L, 
and stopping on the line a d at the point K. Through 
e, and at right angles to e a, draw a line, commenc- 
ing in the point K and terminating in G where it 
meets the line E G. In like manner draw the lines 
of the inner side of the elbow, as shown by F H 
and II I. Draw the miter or joint lines F G, II K 
and L I, as shown. For the patterns proceed as fol- 
lows : Divide the profile into any convenient number 
of equal parts. Place the T-square parallel to E G, 
and, bringing the blade against the points in the pro- 
file, drop corresponding points upon the miter line F 
G. Change the T-square so that its blade shall be 
parallel to the lines of the second section of the elbow, 
and, bringing it against the ooints in F G, cut corre- 
sponding points on II K. Opposite the end of and 
at right angles to the lower arm of the elbow, lay off 
the stretchout line P, as shown, through the divi- 
sions in which draw the usual measuring lines. Place 
the T-square at right angles to the arm of the elbow, 
and, bringing it successively against the points in the 
miter line F G, cut the corresponding measuring lines. 
Then a line traced through the points thus obtained, 
as shown from E to T, will with P constitute 
the pattern of one of the arms. Produce a e, repre- 
senting the middle of the second section in the elbow, 
as shown by Y "W, upon which lay off a stretchout, 
and through the points in the same draw measuring 
lines. Placing the T- s q i,iare parallel to a e, or, what is 
the same, at right angles to the second section of the 
elbow, bring it against the several points in the miter 
lines H K and F G, and cut the corresponding measur- 
ing lines. Then lines traced through the points thus 
obtained, as shown from X to Z and Y to S, will give 
the pattern. 



PROBLEM 42. 
The Patterns for a Five-Piece Elbow. 



To construct tne elevation of a five-piece elbow, 
first draw the profile, as A B C, Fig. 322, from which 
project one of the arms of the elbow, as shown at the 
left by E S R D, continuing its lines indefinitely. At 
right angles to this lay off the other arm, continuing its 
lines till they intersect those of the horizontal arm, or 
till their outer lines intersect, as at g. Draw g a at an 
angle of 45 degrees to either arm, upon which establish 
the point a with reference to the curve which it is de- 



sired the elbow shall have, and from it, at right angles 
to the two arms of the elbow respectively, draw a b and 
a c. From a as center, with a b as radius, describe the 
arc bfedc, which divide into four equal parts, thus 
obtaining the points d, e and/, and draw d a, e a and 
/ a. Then these lines represent center lines of the sev- 
eral sections of which the elbow is composed, at right 
angles to which their sides are to be drawn. 

It may be here remarked that the number of 



134 



Hie New Metal Worker Pattern Book. 



center lines made use of in dividing the quarter circle 
b c represents the number of pieces in the elbow. 
Therefore, to draw an elevation of an elbow in any 
uumber of pieces, construct the quadrant abc as above 
described, then divide I c into such a number of parts 
that the number of lines drawn to a (including a b and 
a c) shall equal the number of pieces required. Thus 
the five lines a b, af, a e, a d and a c are the center 
lines of the five pieces of which the elbow shown in 
Fig. 322 is constructed. Although the two extreme 
lines a b and a c are not, strictly speaking, center lines, 
their relation to the adjacent miter lines is the same as 
mat of the other lines radiating from a. Through/, 
and at right angles to fa, draw V S, joining the side 
of the arm E S in the point S, and joining a corre- 
sponding line drawn through e in the point V. In like 
manner draw the line T R, representing the inner side 
of the same section. The remaining sections are to be 
obtained in the same way. As but one section is 
necessary for use in cutting the patterns, the others 
may or may not be drawn, all at the option of the 
pattern cutter. Draw the miter or joint lines S R, 
V T, etc. Divide the profile (or one-half of it) in the 
usual manner. Place the T-square parallel to the lines 
of the arm, and, bringing the blade against the several 
points in the profile, drop corresponding points upon 
the miter line S R. Shift the T-square so that the 
blade shall be parallel to the part V S R T, and trans- 
fer the points in S R to V T, as shown. For the pat- 
tern of the arm, at right angles to it lay off a stretch- 
out of A B C, as shown by F G, through the points in 
which draw the usual measuring lines. Place the 
T-square at right angles to the arm, and, bringing it 
against the points in R S, cut the corresponding meas- 
uring lines, as shown. Then a line traced through 
these points, as shown from H to I, will be the pattern. 
For the pattern of the piece S V T R prolong the line 
af, as shown by L K, upon which lay off a stretchout, 



through the points in which draw the measuring lines 
in the usual manner. Placing the T-square at right 
angles to S V, or, what is the same, parallel to the 
stretchout line, bring it against the several points in 








If 


/ 1 >. 


1 \ 1 
11 l \ ' . 1 

jl 1 \ 1 / 



Fig. 322 —A Five-Piece Elbow. 

the lines R S and T V, and cut the corresponding 
measuring lines. Then lines traced through the points 
thus obtained, all as shown by 1ST P M, will be the 
pattern sought. 



PROBLEM 43. 

The Patterns for a Pipe Carried Around a Semicircle by Means of Cross Joints. 



In Fig. 323, let F E D be the semicircle around 
which a pipe, of which A C B is a section, is to be 
carried by means of any suitable number of cross 
joints, in this instance ten. Divide the semicircle F E 
D into the same number of equal parts as there are to 
be joints, which, as just stated, is to be ten, all as 



etc., and draw lines from 



shown by D, 0, P, R, S, E 
each point to Z. As there are to be ten joints there 
must necessarily be eleven pieces, therefore, according 
to the directions given in the previous problem, the 
semicircle must be divided into such a number of 
equal parts that the number of lines radiating from Z 



Pattern Problems. 



135 



shall be eleven, all as shown, each line serving as the 
center line of a piece. From D toward the center Z 
set off the diameter of the pipe A B, as shown by the 
point A 1 . From Z as center, with the radius Z A 1 , 
draw the dotted line representing the inner line of the 
pipe, and cutting the radial lines previously drawn in 
the points O 1 , P 1 , etc. Through and O 1 draw lines 
at right angles to Z and continue them in either 
direction till they intersect with the lines drawn 
through P and P 1 on the one side and through D and 
A 1 on the othex. Each pair of lines is to be drawn at 



corresponding to the full sections composing the body 
of the pipe. For the pattern of the end section pro- 
ceed as follows : Divide the profile A C B in the usual 
manner into any convenient number of equal parts, 
and from the points thus obtained carry lines upward 
at right angles to Z D, cutting T' T. Prolong the 
line Z D, and upon it place a stretchout from the 
profile A C B, perpendicular to which draw measur- 
ing lines in the usual manner. With the T-square 
placed parallel to Z D, and brought successively 
against the points in 1" T, cut the measuring lines of 




Fig. SSS.—A Pipe Carried Around a Semicircle by Means of Cross Joints. 



right angles to its respective radial or center line. 
Through the points of intersection draw the lines T T 1 , 
U U', etc., which will represent the lines of the joints 
or miters. 

It will appear by inspection that the point U is 
equidistant from P and 0, and that U' is also equidis- 
tant from P 1 and 1 , and that therefore the lines U IP, 
T T 1 , etc., if continued inward must arrive at the 
center Z. Thus the joint lines, like the center lines, 
must radiate from the center of the semicircle. 

Draw the profile of the pipe A C B directly below 
and in line with one end of the pipe, all as shown in 
the engraving. As may be seen by inspection of the 
diagram, two patterns are required, one corresponding 
to the half section occurring at the end, and the other 



corresponding numbers. Then a line traced through 
the points of intersection thus obtained, as shown by 
I K L, will be the shape of the miter cut, and G I K 
L H will be the complete pattern for one of the end 
pieces. For the pattern of one of the large pieces, 
as U V V 1 IP, lay off a stretchout of A C B upon its 
center line extended, as shown by M 1ST, and through 
the points in it draw measuring lines in the usual man- 
ner. Place the T-square parallel to U Y and, bringing 
it against the points in U IT, cut the line V V. Next 
place the T-square parallel to the stretchout line, 
and, bringing it against the several points in the miter 
lines IT 1 U and V V, cut the corresponding meas- 
uring lines, all as shown, thus completing the pat- 
tern. 



136 



The New Metal Worker Pattern Book. 



PROBLEM 44. 

The Patterns for an Elbow at Any Angle. 



Let D F H K L I G E in Fig. 324 be the eleva- 
tion of a pipe in which elbows are required at special 
angles. In convenient proximity to and in line with 




Fig. 324. — An Elbow at Any Angle. 

one end of the pipe draw a profile, as shown by A B 
C, which divide in the usual manner. Placing the 
J-square parallel to the first section of the pipe, and, 
bringing it against the several points in the profile, 
drop corresponding points upon F G. Shift the 



T-square, placing it parallel to the second section, and, 
bringing it against the several points in F G, drop 
them upon H I. At right angles to the first section 

lay off a stretchout of A B C, 
as shown by T IT, through the 
points in which draw the 
customary measuring lines. 
Placing the J-square at right 
angles to this section of the 
pipe, and bringing it against 
the several points in F G, cut 
the corresponding measuring 
lines. Then the line ELS 
traced through these points 
will, with the line T U, be the 
pattern sought. The pattern 
for the opposite end is to be 
obtained in like manner, all 
as shown by M 1ST P, and 
therefore need not be 
described in detail. 
For the pattern of 
the middle section 
lay off a stretchout, 
W Y, at right angles 
to it, with the cus- 
tomary measuring 
lines. Placing the 
T-square at right 
angles to the section, bring it successively against 
the points in G F and I H, and cut the correspond- 
ing measuring lines, as shown. Then lines traced 
through these points, as shown by Y X Z Q, will 
be the pattern sought. The positions of the longi- 
tudinal joints in the several sections of this elbow, as 
well as those of all others, are determined by the 
order in which the measuring lines drawn through 
the stretchout are numbered. In the present instance 
the joints are allowed to come on the back of the 
pipe, or, in other words, upon D F H K, which 
corresponds to the point 1 in the profile. Hence, 
in numbering the measuring lines in the several stretch- 
outs, point I is placed at the commencement and 
ending, while if it were desired to have the joint 
in either piece come on the opposite side, or at a point 
corresponding to 9 of the profile, the stretchout would 
have commenced and ended with that figure, the 



Pattern Problems. 



13? 



figure 1 in that case coming, in regular order, where 9 
now occurs. The effect of such a change upon any of 
the patterns here given would, be the same as if they 



were cut in two upon the line 9 and the two halves 
were transposed. 



PROBLEM 45. 

The Patterns for a Bifurcated Pipe> the Two Arms Being: the Same Diameter as the Main Pipe, and Leaving; It at 

the Same Angle. 



In Fig. 325 is shown an elevation of a bifurcated 
pipe, all arms being of the same diameter. In this 
problem, as in many others, it becomes necessary to 
first make a correct drawing of the intersection of the 
parts showing the miter lines correctly; after which 



of the miter lines will at once be determined. Thus 
the intersection of the three bisecting lines at E gives 
the point at which the miter lines starting from the 
points P, E and K must meet. 

In line with the upper end of the pipe draw a 




Fig. 325. — The Pattern for a Bifurcated Pipe. 



the method of laying out of the miter patterns is the 
same as that employed in several other problems 
immediately preceding this. If, in this case, each arm 
of the pipe be divided longitudinally into two equal 
parts, as shown by the center lines, and each half be 
considered as a separate molding the correct position 



profile of it, as shown by A C B. A profile will also 
be needed in one of the oblique arms, a half only 
being shown at A' C B' on account of the limited 
space. For the pattern of the upper portion of the 
pipe, divide the profile A C B into any number of 
equal spaces, and place the stretchout of the same on 



138 



The New Metal Worker Pattern Book. 



any line drawn at right angles to S P, as shown by 
the continuation of S D to the left, and draw the 
usual measuring lines. Next drop the points from 
the profile A C B parallel with S P till they cut 
the miter line PRE; then placing the T-square at 
right angles to S P, drop the points from the miter 
line P E into measuring lines of corresponding num- 
ber. A line traced through these points of intersection, 
as shown from E" to P', will give the miter cut on the 
lower end of the pipe S D E P, one-half of which only 
is shown in the engraving. The pattern for the piece 
E F J K is obtained in exactly the same manner, and 
might be obtained, so far as the half indicated 



by C B' on profile is concerned, from the original 
profile, by simply continuing the lines through to the 
miter line J F, as shown. For simplicity, therefore, 
the profile A' C B' is divided into the same number 
of equal parts as the original profile, and a stretchout of 
it is placed upon any line, as T IT, drawn at right angles 
to E F. The points are then dropped from the profile 
both ways, cutting the miter lines KEE and J F ? 
after which, with the T-square placed parallel to T TJ r 
they can be dropped into the measuring lines of the 
stretchout. Lines traced through the points of inter- 
section -will constitute the required pattern, as shown 
by K' E' E' R" K" X W V. 



PROBLEM 46. 



The Patterns for the Top and Bottom of a "Common" Skylight Bar. 



In Fig. 326, A B represents a portion of the pro- 
file of the ridge bar, or of the ventilator forming the 
top finish of a skylight, against which the upper end 
of a " common ' ' bar is required to miter ; and C D 
represents the profile of the curb or finish against 
which the lower end of the bar miters. The parallel 
oblique lines connecting the two show the side eleva- 
tion of the bar whose profile is shown at E F F 1 . 

As the profile consists of two symmetrical halves, 
either half, as E F or E F 1 , may be chosen to work 
from, and as it contains no curved portions it is 
simply necessary to number all of its points or angles, 
and then to place a complete stretchout of the same 
upon any line drawn at right angles to the lines of the 
molding, as G H, and to draw the usual measuring 
lines, all as shown. As a properly drawn elevation 
shows the intersection of the points of the profile with 
the two miter lines A B and C D 1 , it is only neces- 
sary to place the T" s q uare parallel to the stretchout 
lines G H, and bring it successively against the points 
in A B and C D', and cut corresponding measuring 
lines, as shown at I J and K L. Straight lines con- 
necting the points of intersection will complete the 
pattern, as shown at I J L K. The length of the 
pattern, which is here shown indefinite, must be 
determined by a detail drawing, in which the rise M 
B and the run M D' are conectly given. 

The patterns for the ' ' jack ' ' bar and for the 



" hip " bar will be given later among those problems 

in which the development of the miter line and the 

J/ 




Fig. SS6.—The Patterns for a " Common " Skylight Bar. 

raking of the profile are necessary, with which they 
are properly classed. 



Pattern Problems. 



139 



PROBLEM 47. 



The Patterns for a T-Joint Between Pipes of the Same Diameters. 



Let D F G H M I K E in Fig. 327 be the eleva- 
tion of two pipes of the same size meeting at right 



B' 4 




Fig. S27.—A T-Joint between Pipes of the Same Diameter. 

angles and forming a T, of which ABCD and A 1 B' 
C D' are profiles drawn in line with either piece. As 
the two profiles are alike, and as the end of one piece 
(D E F K) comes against the side of the other piece 
(G I M H), both halves of D E F K, B A D and B C 



D, will miter with one-half, B 1 C D', of the piece G I 
M H. By projecting the points B and B' from the 
profiles through their respective elevations the point L 
is found, which being connected with the points F and 
K gives the miter lines. Space the profile ABCD 
into any number of equal parts and lay off the stretch- 
out N at right angles to the pipe of which ABC 

Dis the profile, as shown, 
through the points in 
which draw the usual 
measuring lines. Set the 
T-square at right angles 
to this pipe, and, bring- 
ing the blade against 
the several points on 
the miter lines, cut the 
corresponding measuring 
lines drawn through the 
stretchout, as indicated 
by the clotted lines. Then 
NF'UTWO will be 
the pattern for the upper piece. As both halves of 
this piece (dividing now upon the line A C) will be 
alike only one-half of the profile (A B C) has been 
divided, but the stretchout is made complete. For 
the pattern of the other piece, divide its profile into 
any convenient number of equal parts and lay off 
the stretchout on the line B T, drawn at right angles 
to the pipe. Placing the T-square parallel with 
the pipe drop points upon the miter lines from that 
portion of the profile (B 1 C D") which comes in line 
with them ; then place the blade of the T-square at 
right angles to the pipe, and, bringing it against the 
several points in the miter lines, cut the corresponding 
measuring lines, as shown by the dotted lines. A 
line, X Y Q Z, traced through these points will 
bound the opening to be cut in the pattern for the 
lower pipe. From the points 1 in the stretchout 
draw the lines R P and T S, in length equal to the 
length of the pipe. Connect P S. Then PETS 
will be the required pattern. The seam in the pipe 
may be located as shown in the engraving, or at some 
other point, at pleasure. 



140 



The New Metal Worker Pattern Book. 

PROBLEM 48. 



The Patterns for a Square Pipe Describing a Twist or Compound Curve. 



As problems of this nature frequently occur in 
connection with hot air pipes, grain chutes, etc., 
this problem is given as embodying principles which 
can often be made use of. The upper opening of the 
pipe in this case is required to be in a horizontal 
plane, while the lower opening is in a vertical position 
and placed at a given distance below and to one side 
of the top, the pipe describing a quarter turn when 
viewed from either the top or the front. 

To more fully illustrate the nature of the prob- 
lem, a perspective view of it is shown in Fig. 328, in 
which the pipe is represented as being contained 
within a cubical shaped solid. The solid, of which 
the pipe is represented as forming a part, is shown in 
outline, the pipe itself being shaded to show its form, 
while upon the front and lower side of the solid are 
shown in dotted lines the front elevation and plan of 
the pipe. Thus G F T C represents the front view of 
a solid just large enough to contain the pipe, in which 
A B C D shows the position of the lower opening, and 
A B E F D C shows the curve of the pipe as seen from 
the front. Gr H S F is the top of the solid in which 
the upper opening N P S E is situated. The curve of 
the pipe in plan has been projected to the lower face 
of the solid by vertical lines, B L, and others not 
shown, and is shown by C J K L M D. To state the 
case simply, then, A B E is the profile of the piece of 
metal forming the top of the pipe, while DML and 
C J K are the two miter lines, or the plans of the 
intersecting surfaces, and C D F is the profile of the 
lower side of the pipe intersecting the same miter 
lines. The top and bottom pieces being developed, 
it is only necessary to reverse the operation and con- 
sider the lines of the plan DML and C J K as the 
profiles of the front and back pieces respectively, 
while ABE and CDF become the miter lines or 
elevations of the intersecting surfaces. 

A part of these operations are carried out in 
detail in Fig. 339, where the elevation and the plan 
are drawn directly in line with each other; the various 
points being represented by the same letters in the 
two illustrations. For the pattern of the top piece 
divide its profile A B E by any convenient number of 
points (1, 2, 3, etc.), from which drop lines ver- 



tically cutting the two miter lines D' M and C J of 
the plan, as shown (the figures of the plan 2 to 11 
have no reference to this part of the operation). Upon 
B 8, drawn at right angles to the direction of the mold, 
lay off the stretchout of ABE, through which draw 
the usual measuring lines. With the J-square placed 
parallel to B 8 and brought against the several points 
in the two miter lines cut lines of corresponding num- 
ber ; lines traced through the points of intersection, as 
shown by E T and U S, will give the pattern of the top 
piece. It will be noticed that owing to the contrary 
relation of the two curves it is necessary to have the 
points of the profile occur more frequently near B 
than E, as otherwise they would intersect the miter 




Fig. 828. — Perspective View of a Pipe Describing a 
Twist or Compound Curve. 



line D' M too far apart near D', while they would 
occur more frequently than is necessary near M. As 
there is no curve from A to B of the profile, that part 
of the pattern from B to S will be a duplicate of the 
plan view, consequently the curve from B to the 
measuring line drawn from S may be traced from the 
plan. The development of the pattern for the lower 



Pattern Problems. 



141 



side of the pipe is not given, but it would be accom- 
plished in exactly the same manner as that of the top 
piece, using CDFas the profile instead of A B E. 

For the pattern of the front piece of the pipe, 
divide its profile L M D' by any convenient number 



the T-square placed parallel to Q 1 and brought against 
the various points in B E and D F cut corresponding 
measuring lines. Lines traced through the points of 
intersection, as shown by Q P and N, will give the 
required pattern. 




S 1 2 3 4 5 6 7 



Fig. 329. — Patterns for a Pipe Describing a Compound Curve. 



of points 1, 2, 3, 4, etc., from which drop lines ver- 
tically, cutting the two miter lines D F and B E, as 
shown. Upon Q 1, drawn at right angles to the direc- 
tion of the mold, place the stretchout of L M D', 
through which draw the usual measuring lines. With 



The pattern of the back piece not given in the 
illustration can be developed in exactly the same man- 
ner as that of the front by using C J K as the profile 
and proceeding otherwise the same as in the fore- 
going. 



142 



The New Metal Worker Pattern Book. 



PROBLEM 49. 



The Construction of a Volute for a Capital* 



It is sometimes desirable in designing capitals of 
large size to construct the volutes of the same of strips 
of metal cut and soldered together. The principal 
characteristic entering into the design of the volute, 
and that which distinguishes it from an ordinary scroll, 
consists in a pulling out or raising up of each succes- 
sive revolution of the scroll beyond the former, thus 
producing a ram's horn effect. This feature of its de- 
sign is also frequently embodied in the construction of 
scrolls used to finish the sides of large brackets or head 
blocks, such as may be seen by reference to Fig. 87 on 
page 12. As all volutes, except those of the Ionic 
order, always occur under the corners of the abacus 
and project diagonally from the bell of the capital, 
their forms can only be correctly delineated in a 
diagonal elevation. 

In Fig. 330 is shown a diagonal elevation of a 
portion of the bell and abacus of a capital with the 
volute. Immediately below the same, DACB shows 
one-quarter of the plan of the capital, turned to corre- 
spond with the elevation, in which the various curves 
of the volute have been carefully projected from the 
elevation, as shown by the dotted lines. As the pat- 
tern cutter is dependent upon the drawing of the plan 
for his miter lines, considerable care must be given to 
this part of the work. On account of the small scale 
necessary in drawing Fig. 330, an enlarged view of the 
plan of the helix of the volute, as seen from below, is 
shown in Fig. 331, in which the various curves can be 
followed throughout their course. 

The volute as here given consists of two side 
pieces or scrolls, an outside cover or face strip, an in- 
side cover and two narrow strips to fill the space where 
the second curve of the scroll projects beyond the first. 
The outside cover or face strip extends from F of the 
elevation to G, where it is met by the inside face strip, 
which begins at H. To obtain the pattern for the in- 
side cover, divide the profile from H to Gr into any 
convenient number of equal spaces, and lay off a 
stretchout of the same upon the center line of the 
volute in plan, A B, extended toward K, as shown by 



the nine spaces on tne upper side of the line. Drop 
lines vertically from each of these points intersecting 
the upper line of the side of the scroll in plan. Place 
the T-square parallel to the stretchout line B K, and 
bringing it successively against the points in the plan, 
drop lines cutting corresponding lines of the stretch- 
out. Then a line traced through the points of inter- 
section, as shown from I to J, wdl give the shape of 
the side of the strip to cover the space between the 
points H and Gr of the elevation. A similar course is 
to be pursued in obtaining the outside cover or strip 
extending from F to Gr. This stretchout consists of 
fourteen spaces, and is shown on the lower side of the 
center line A K, the pattern being shown from L to 
M. The pattern for the remaining strip consists of a 
stretchout of seven pieces taken from the profile be- 
tween G and the termination of the scroll line. Points 
from this part of the profile are intersected with two 
miter lines in the plan, one forming the outer line of 
the strip, or its finish against the more projecting part 
of the scroll, and the other forming its finish against 
the lower scroll or inner edge of its first or outer curve. 
In Fig. 331 the lines showing the projection of the 
inner part of the volute beyond the outer curve are 
clearly seen. In the lower half the lines correspond- 
ing to the points 1 to 7 of the profile are shown by cor- 
responding numbers. Lines dropped from the points on 
both these lines to corresponding lines of the stretchout 
will give the pattern as shown from M to N. 

By inspection of the drawing it will be seen that 
the outline of the volute, as given in the elevation, 
does not represent exactly the "true face" of the 
scroll. As the variations in the angle of the side of the 
central part or helix of the scroll are only such as can be 
produced by the springing of the metal necessary to 
bring it into shape, no allowance need be made for such 
variation in cutting the pattern directly from the eleva- 
tion. Careful measurements of the stem or lower 
part of the volute, as shown in the plan, however, show 
that the distances from point 9 to points a and b, if laid 
off on a line parallel to A B, would reach to points a 2 



Pattern Problems. 



U5 



and b\ These points projected back into the elevation 
locate them in that view at a and b\ Therefore the 
outline of the back of the stem will have to be ex- 
tended as shown by the dotted line from F to a % . This 



and need not be repeated here. The correct outline, from 
G to b 3 is omitted to avoid confusion with the figures. 

To avoid confusion of lines in dropping the points 
from the different parts of the profile to the miter lines 




PATTERNS 
Fig. 330,— The Construction of a Volute for a Capital. 



outline can be accurately obtained, if deemed necessary, 
by the raking process described in connection with a 
number of other problems in this section of this chapter, 



and thence to the stretchout, only the first and last of 
each series or stretchout have been shown by dotted 



lines in the drawing. 



144 



The Neiu Metal Worker Pattern Book. 

PROBLEM 50. 



The Pattern for a Pyramidal Flange to Fit Against the Sides of a Round Pipe Which Passes Through Its Apex. 



A pictorial illustration of the flange fitting against 
the sides of the pipe, as stated above, is shown in Fig. 
832. In Fig. 333 K L M represents the elevation of 




Ml | 

III 

III 


III 


III 


III 1 


III 1 


III | 




Fig. S3S. — Pattern for a Pyramidal Flange to Fit Against a 
Round Pipe. 

pyramid, and PRTS elevation of the pipe that is to 
pass through it, A B D C being plan of pipe and pyra- 
mid. As the pyramid has four sides, each side will 
miter or fit against one-quarter of the profile of the 
pipe, as will be seen by reference to the plan. Again, 
as each side consists of two symmetrical halves, as 
shown by the dotted line dividing the side B D, one- 
eighth of the profile of the pipe (as Gr I) is all that need 



be used in obtaining the pattern. Therefore, divide 
Gr I into any convenient number of parts and carry ver- 
tical lines to L M, which represents one side of the 
pyramid, and then, from these points and 
the points L and M carry lines at right 
angles to it indefinitely, as shown. L M 
in the elevation represents the complete 
length of one side of the pyramid, as it 
would be if not cut by the pipe. Lay off 
on the line from M the length of one side 
of the base of the pyramid, as B D in the 
plan, as shown by M M 1 . Bisect M M' 
at F, from which point draw F E parallel 
to L M, cutting the line from point L at 
E. The lines from E to the points M 
and M 1 would give the pattern of one side of the 
pyramid if it were not to be cut by the pipe. It 
simply remains now to measure the width of the pat 
tern at the various points ofjthe curved portion, which 




Fig. 832.— Perspective View of Pyramidal Flange. 

can be done by measuring the distance of each point in 
the profile Gr I, from the center line of the side B D in 
plan, and setting off these distances upon lines of 
corresponding number drawn through the pattern from 
the line L M, measuring each time from the center line 
E F. Thus the distance of point 4 from the center 
line in plan is set off from the center line of the pat- 
tern each way upon line 4, and coincides with this 
point as previously established by the lines drawn from 
E to M and M 1 . The distance of the point 3 from the 
center line in the plan is set off from the center line of 
pattern each way upon line 3 of the pattern. Point 2 
is established in the same manner. A line traced 
through the points 4 3 2 12 3 4 completes the pat- 
tern. 



Pattern Problems, 

PROBLEM 51. 



145 



The Patterns for a Square Pyramid to Fit Against the Sides of an Elliptical Pipe Which Passes Through Its Center, 



In Fig. 334, A B C D shows the plan of a square 
pyramid, whose apex, if completed, would be at E. 
F H I J shows the horizontal section of an elliptical 
pipe, against the sides of which the sides of the pyra- 
mid are required to be fitted. From the side A B (or 



patterns will be necessary, one for A B Gr S to fit 
against the broad side of the pipe, and another for 
B G T C to fit against what might be termed the 
edge or narrow side of the pipe. To obtain the 
pattern of the side of the pyramid shown by B Gr T 




D c c 2 

Fig. 334. — The Patterns for a Square Pyramid to Fit Against the Sides of an Elliptical Pipe, 



D C) of the plan is projected a front elevation, in which 
KLMN shows the broad side of the pipe. To the 
right another or side elevation is projected, in which 
the narrow view of the pipe is shown by OPQE. 
An inspection of the plan will show at once that two 



(or B E C, if the pyramid were complete), first 
divide that portion of the profile of the pipe from 
Gr to H by any convenient number of points, as shown 
by the small figures, from which, together with B and 
E, project lines vertically to the elevation above, cut- 



146 



The New Metal Worker Pattern Book. 



ting that side in profile as shown from E 1 to B\ At 
right angles to E 1 B 1 carry lines from each of the points 
indefinitely, as shown. At any convenient distance 
•away, cut these lines by any line, as E 3 V 3 , drawn 
parallel to E 1 B 1 . Upon each of the lines drawn from 
the points in E 1 B 1 , set off from E 3 V s the distances 
oxpon lines of corresponding number in the plan meas- 
ured from E Y. Thus upon line 5 of the pattern set 
■off either way from its intersection with the line E 3 V s 
a length equal to the distance of point 5 of the plan 
from the line E V. Upon line 4 of the pattern set off 
distances equal to that of point 4 from E V of the plan, 
etc. Also make V s C 3 and V 3 B 3 equal to V C and 
V B of the plan. Lines drawn from C 3 and B 3 toward 
E 3 will meet the points previously set off on line 5 of 
the pattern, indicated by T 1 and G 1 , and will constitute 
the sides or hips of the pattern, and a line traced 
through the points set off on lines 1 and 5 inclusive 
will give the shape of that portion of the pattern to fit 
against the pipe. 

An exactly similar course is to be pursued in ob- 
taining the pattern of the side of the pyramid ASGB 



(or A E B of the complete pyramid), whose profile is 
shown by B 2 E 2 of the side elevation, showing the nar- 
row view of the pipe. The pattern is shown at A' B 4 
G 3 S 2 , and the operation is clearly indicated by the 
lines oi projection. 

If it is desired to complete the elevations by show- 
ing the lines of intersection of the sides of the pipe 
with the sides of the pyramid shown respectively in 
each elevation, as from c to b and e to/, it can be ac- 
complished as follows : To obtain the line c b, erect lines 
vertically from points 6, 7 and 8 (not shown), passing 
through the space between c and b in the front elevation, 
upon each of which set off the hight of each point as 
measured upon lines of corresponding number from B 2 
C 2 to B 2 E 2 , as shown from B toward e, in the side eleva- 
tion ; then a line traced through the points thus obtained 
will give the line c b. In the same manner lines from the 
points 1,2,3 and 4 can be carried at right angles to B 2 C 2 
into the side elevation, upon which to set off hights of 
corresponding points as measured from A' B' to B' E', 
as shown between a and b in the front elevation ; then 
a line traced through the points will give the line ef. 



PROBLEM 52. 



The Patterns tor a Rectangular Pipe Intersecting: a Cylinder Obliquely 



In Fig. 335, let A B C represent the plan of a 
drum or cylinder, and BEDC the plan of rectangular 
pipe, the profile of which is shown by F G II I. In 
the elevation, J K L M represents the drum, NOPQ 
the rectangular pipe, and K n the angle at which 
they are to intersect. Draw the end view, or plan, of 
circular drum in line with the elevation, as shown. 
Also extend n and P a so a line dropped from point 
C of plan will cut them, as shown by N and Q. Then 
n 1ST Q a is the joint between the drum and pfpe, as 
shown in elevation. For the pattern of rectangular 
pipe proceed as follows : Divide B C of plan into any 
convenient number of equal parts, and from these 
points carry lines horizontally cutting E D. Also from 
the points in C B drop lines vertically cutting Q P and 
N 0. On P extended lay off a stretchout of profile 
F G H I, as shown by I I', transferring the spaces in 
E D to H G and F I', and through the points in it draw 
the usual measuring lines, as shown. Place the J-square 



parallel with the stretchout line 1 1', and, bringing it 
successively against the points in the miter lines N n 
Q q, cut the corresponding measuring lines, as indicated 
by the dotted lines. Lines traced through the points 
thus obtained, as indicated by ihgfi', will give the 
desired pattern. It will be observed that I H h i is a 
duplicate of P Q N, and that GF/^ is also a dupli- 
cate, only in a reversed position. The points in hg of 
pattern are derived from Q q, as the points in / i' of 
pattern are derived from IS!" n. If the size of the work 
is such as to render it inconvenient to drop points 
from the elevation to the pattern by means of the T- 
square, the stretchout line I I' can be drawn where 
convenient, the usual measuring lines erected and the 
distances from P to points in 1ST n and Q q transferred 
by means of the dividers to lines of similar number 
drawn from the stretchout line. 

For the pattern or shape of opening in drum, pro- 
ceed as follows : On L M extended, as E U, lay off a 



Pattern Problems. 



147 



stretchout of B C of plan, and from the points thus ob- 
tained erect the usual measuring lines, as shown. Place 



ing lines of corresponding number. Through the 
points thus obtained trace the lines V W and Y X ; 



PLAN 




Fig. SS5.— Patterns for Rectangular Pipe Intersecting a Cylinder Obliquely. 



l i< 



the T-square parallel with M L, and, bringing it suc- 
cessively against the points 1ST n and Q q, cut measur- 



then V W X Y will be the shape of the required open- 
ing; in the side of the drum. 



PROBLEM 53. 

The Pattern lor the Intermediate Piece of a Double Elbow Joining: Two Other Pieces Not Lying- in the Same Plane. 



In Fig. 336 is shown a front and side view of a 
somewhat complicated arrangement of elbows such as 
sometimes occurs when pipes have to be carried around 
beams or through limited openings. An inspection of 



the drawing will show that once the correct angle of 
the different elbows is ascertained the development of 
the miters will be quite simple, and is the same as 
those occurring in several of the problems preceding 



148 



Tlxe New Metal Worker Pattern Book. 



this. The lower section of the pipe rises vertically to 
the first elbow, B, from which it must be carried up- 
ward a distance equal to C M, to the left a distance 
equal to B M, as shown in the front view, and back a 
distance equal to o C, as shown by the side view. 



methods employed in drawing the two views shown in 
Fig. 336 will be of assistance to the pattern cutter. 
According to the principles of projection each indi- 
vidual point must appear at the same hight in both ele- 
vations, and at the same distance right or left and for- 




FRONT VIEW 



SIDE VIEW 



Fig. 330.— Elevations of Double Elbow. 





D 

K 

i \ 



1C 



Fig. 338.— Diagram Used in Obtaining 
Correct Side View of Upper Eibow, 



Fig. 337.— Correct Side View of 
Lower Elbow. 




1 2 3 



Fig. 340.— Method of Obtaining the Pattern of Middle 
Portion in One Piece. 



Fig. 339.— Correct Side View of Upper Elbow. 
The Pattern for the Intermediate Piece of a Double Elbow Joining Two Other Pieces Nut Lying in the Same Plane. 



From the elbow C it then rises vertically, as seen in 
front, but really toward the observer as shown by the 
side view. The problem then really consists in find- 
ing the correct angles of the elbows, and becomes a 
question of draftsma.nship rather than of pattern cut- 
ting. Some suggestions then with regard to the 



ward or back, with reference to the center lines of the 
plan. As front and side views are here required, 
begin by first placing the given plan in two positions, 
turning those sides of it to the bottom which corre- 
spond to the sides required in the elevations, and 
proceed by erecting the center lines of the differ- 



Pattern Problems. 



149 



ent pieces in their proper positions and building the 
pipe around them, so to speak. The plan being 
a circle, the different sides can only be indicated by 
numbering the joints, as will be seen by referring 
to the plans, point 2 appearing in front in the front 
elevation, and point 3 appearing in front in the side 
elevation. The plans having been so arranged and 
corresponding parts in both given the same number, 
proceed now to erect the center line of the lower sec- 
tion, making the hight of the first bend, B, the same in 
both views, as indicated by the dotted horizontal line. 
From this point the center line is continued in both 
views, giving it its proper inclination to the left in the 
front view, and to the right in the side view, all accord- 
ing to the specified requirements, thus establishing the 
point C, making it agree in hight in both views. From 
this point the pipe appears inclined only in the side 
view, which means that it leans toward the observer 
in the front view. Next draw the outlines of the pipe 
at equal distances from the center line and on either 
side of it throughout the entire course of the pipe in 
both views, deriving them from the points of plans 1 
and 3 in the front view and 2 and 4 in the side view. 
Their intersection in the front view will give definitely 
the positions in the miter of points 1', l 2 , and 3', 3 1 , 
and in the side view of points 2', 2 3 , and 4', 4 a . As 
point 3' has been established in the front view, if a line 
be carried horizontally across till it intersects the line 
from point 3 of the side view, it will give the hight of 
point 3' in the miter, as shown in the front view. In the 
same manner a horizontal line from 1' in front, inter- 
secting the perpendicular from point 1 in plan of side, 
will give the true hight of point 1' in the side view. 
A careful inspection of the dotted lines of Fig. 336 will 
make the subsequent operations necessary to the com- 
pletion of the elevations clear to the reader without 
further explanation. Since neither of the views gives 
a true side view of the intermediate piece, one must 
be constructed from the facts now known, so as to get 
the true angle of the elbow B. By dropping a vertical 
line from the point C of the front view into the plan it 
will appear that the horizontal distance between the 
points C and B would be measured by the line E P of 
the plan ; but by further reference to the side eleva- 
tion the position of the point C is found to be to the 
right of its center line by a distance equal to B C 
of the plan ; therefore, if this distance be set off on the 
vertical line from the point E in the plan below the 
front view, which is indicated by E C, the point C will 
•determine the true position in the plan of the point 



C of the elevations, and the distance C P will be its 
horizontal distance from B. Since, now, its vertical 
distance can easily be obtained from either front or side 
elevation, a new diagram can now be easily constructed 
which shall contain the proper dimensions to obtain a 
correct side view of this elbow. Proceed, then, to 
construct diagrams shown in Fig. 337, making C M 
equal to C M, Fig. 336, M B equal to C P of the plan, 
Fig. 336 ; a line connecting the points C and B will 
represent the center line of the intermediate portion of 
the pipe and give its true relation to the vertical por- 
tion whose center line is represented by B H, Fig. 
337. By drawing the outlines of the pipe at the re- 
quired distance on either side of the center lines B H 
and B C, a correct side view of the miter is obtained. 
Since, as has been referred to above, the tipper portion 
of the pipe appears vertical in one view and inclined in 
the other (see Fig. 336), a correct side view of the 
upper elbow is more difficult to be obtained. While 
different methods may be devised for obtaining it, the 
following is perhaps the simplest : As the upper sec- 
tion of the pipe, as shown by Fig. 336, is of indefinite 
length, any point may be assumed, as D, from which 
to take measurement for obtaining the angle of the 
upper elbow. Since the true length of the line C B of 
either elevation has already been obtained and given 
in Fig. 337, and since the true length of the part C D 
can be derived from the side view of Fig. 336, it is 
necessary only to obtain the true distance between the 
points D and B of the elevations to obtain the proper 
angle at the point C. By dropping a vertical line from 
the point D to a horizontal line drawn from the point C 
in the side view, Fig. 336, the horizontal distance be- 
tween C and D may be obtained. By tranferring this 
distance, o C, to the plan of the front view, and locat- 
ing its distance from C, as indicated by D, this point 
will give the true position of the point D in the plan, 
and the line D P will give the true horizontal distance 
between the points D and B. In Fig. 338 let the 
distance C be equal to the line D P of Fig. 336. At 
point erect a perpendicular, D, making the dis- 
tance O D equal to o D of the side elevation, Fig. 336. 
From the point C drop a perpendicular, C B, making 
that distance equal to the vertical hight between the 
points C and B, as measured on line C M of the front 
view ; a diagonal line connecting the points D and B 
will readily be seen to give the true distance between 
the points bearing those letters in Fig. 336 Proceed 
now to construct the triangle shown in Fig. 339, mak- 
ing C B equal to C B of Fig. 337. From C as a center, 



150 



Tlie New Metal Worker Pattern Book. 



with a radius equal to C D as obtained from the side 
view in Fig. 336, draw a small arc, which intersect 
with the arc drawn from the point B, with a radius 
equal to B D as obtained in Fig. 338 ; this will give the 
correct angle of the upper elbow at C. A complete 
view of the miter may be obtained by further adding 
outlines of the pipe at equal distances on either side of 
the center lines, and connecting their angles, as shown 
by the line b f. Having now obtained two correct side 
views of the two elbows, the problem of obtaining the 
patterns for the same can be solved by the regular 
method. 

To obtain the pattern for the middle portion in 
one piece further calculations, however, will be re- 
quired. This, of course, could be obviated by making 
a slip joint in the middle portion of the pipe, by means 
of which the two elbows could be made separate, and 
then simply turned upon each other till the required 
angle is obtained. But as it might be desirable to 
make the pattern of the middle jDortion in one piece some 
means must be employed of ascertaining just how far 
one elbow would have to be turned upon the other 
were they made separately. As the seam in pipe con- 
taining elbows is usually made at either the shortest or 
the longest point of the miter, it may be easily seen, by 
an inspection of Fig. 336, that a line from the shortest 
point, or throat, b of the upper miter of the piece in 
question, would not meet the longest point, or point a, 
Fis'. 337, in the miter of the other end, and some 
means must be devised for obtaining the real position 
of these points, of which the following is perhaps the 
simplest : From either of the points D or B, Fig. 339, 
draw a line through the point b, continuing it to the 
further side of the triangle, as indicated by the line 
B X. Lay off the distance D X upon the line D C of 
the side view, Fig. 336, thereby locating the position 
of the point X in that view. Aline connecting this 
point with point B must intersect the miter line 2 2 , 4 = , 
in this view at the same point which it does in Fig. 
339, thereby locating its position just as much as 
in Fig. 339. This point having been obtained, its 
equivalent upon the lower miter may be found by 



means of a line drawn parallel to the center line of the 
middle portion, intersecting it at the point g, from 
which point it can be carried vertically to the plan, as 
shown by Z, where its distance from other points can 
be measured with accuracy. The position of the point 
a in Fig. 337 will readily be seen to be at point h in 
the plan of the front view, Fig. 336. By transferring 
the point Z from the plan of the side view to the plan 
of the front view, which can be done by measuring its 
distance from either of the points 2 or 3, the t'elative 
position of the points h and Z upon the same circle will 
be apparent. Fig. 340 shows a diagram, in which a 
correct side view of the two elbows is shown, giving 
the proper distance between the points B and C. 
Considering the lower one to be in its proper and fixed 
position, the profile is constructed and divided into 
points for the purpose of obtaining a stretchout and 
the miter pattern according to the usual method, the 
stretchout being shown upon the line E F in the 
profile and point 8 will readily be seen to correspond 
with point h in the plan of the front view. The posi- 
tion of the point Z in the same plan can be obtained 
by measuring its distance from point h and transferring 
it to Fig. 340, as indicated by M. As the point b of 
the upper elbow is in relation to the highest point, or 
a, of the lower elbow as the point M is to the point 8 
in the profile, it becomes necessary to place the point 
S in the stretchout of the upper elbow as far from the 
point 8 on the stretchout of the lower one as the dis- 
tance from 8 to M in the profile, which is shown by 
m in the stretchout. The stretchout of the upper 
elbow is thus moved, as it were, in its relation to the 
stretchout of the lower elbow, that portion of it which 
extends beyond the point 1 at the left end being added 
to the other so as to make the seam continuous. The 
points are then dropped from the profile to the two 
miter lines, and thence into measuring lines of corre- 
sponding number in the stretchout. Lines traced 
through the points of intersection, as shown byYPR 
X, will be the required pattern. The miters for the 
upper and lower sections would, of course, be inverted 
duplicates of the adjacent ends of the middle piece. 



Pattern Problems. 



151 



PROBLEM 54. 

A Joint Between Two Pipes of the Same Diameter at Other Than Right Angles. 



Let L F D E K I H M of Fig. 341 represent the 
elevation of two pipes of the same diameter meeting at 
the angle M H I, for which patterns are required. 
Draw the profile or section A' B 1 C 1 in 
line with the branch pipe, and the section 
A B C in line with the main pipe. As 
hoth pipes are of the same diameter, and 
the end of one piece comes against the 
side of the other piece, both halves of 
the branch pipe (dividing at the point B) 
will miter with one-half, B D, of the main 
pipe. By projecting lines through the 
elevations of each piece from the points B 
or 4 of their respective profiles the point 
G is obtained, which, being connected with 
points F and H, gives the required miter 
line. Space both the profiles into the 
same number of equal divisions, com- 
mencing at the same point in each. For 
the pattern of the arm proceed as follows : 
Lay off the stretchout O N opposite the 
end of the arm and draw the usual meas- 
uring lines at right angles through it, as 
shown. Place the J- square at right 
angles with the arm, or, what is the same, 
parallel with the stretchout line, and, 
bringing the blade successively against the 
points in the miter line F G H, cut the corresponding 
measuring lines. Through the points thus obtained 
trace the line PEST, which will form the end of the 
pattern required. For the pattern of the main pipe 
proceed as follows : Opposite one end lay off the 
stretchout, as shown by V Y, and opposite the other 
end lay off a corresponding line, as shown by U X. 
Connect U V and X Y. From so many of the points 
in the stretchout line V Y as represent points in the 
half of the profile BAD draw the usual measuring 
lines. With the T-sqaare placed parallel to the mold- 
ing D I, drop the points from the profile onto the 
miter line F G H ; then, placing it at right angles to 
the molding, drop lines from the points in the miter 
line intersecting the corresponding measuring lines. 
A line traced through these points of intersection, as 
F 1 Z H 1 W, will describe the shape required. The 
position of the seam in both the arm and the main pipe 
is determined by the manner of numbering the spaces 



in the stretchout. In the illustration the seam in the 
arm is located in the shortest part, or at a point corre- 
sponding to 1 of the profile. Accordingly, in number- 




XL 



— MY 



Fig. 241.— A Joint between Two Pipes of the Same Diameter at 
Other than Right Angles. 



ing the divisions of the stretchout, that number is 
placed first. In like manner the 'seam in the main pip© 
is located at a point opposite the arm. Therefore, in 



152 



The New Metal Worker Pattern Book, 



numbering the spaces in the stretchout commence 
at 1, which, as will be seen by the profile, represents 
the part named. If it were desirable to make the seam 
come on the opposite side of the main pipe from where 
it has been located — that is, come directly through the 
opening made to receive the arm — the numbering of the 
stretchout would have been begun with 7. In that case 
the opening FWH'Z would appear in two halves, and 
the shape of the pattern would be as though the pres- 
ent pattern were cut in two on the line 7 and the two 
pieces were joined together on lines 1„ By this expla- 
nation it will be seen that the seams may be located 



during the operation of describing the pattern wherever 
desired. It is not necessary, as prescribed at the out- 
set of this problem, that both profiles should be spaced 
off exactly alike. Any set of spaces will answer quite 
as well, provided there be points in each exactly half 
way between A and B of either profile — that is, where 
points 4 are now located. They are spaced alike in this 
case to show that lines dropped from points of the same 
number in each profile arrive at the same point on the 
miter line, and that therefore when both pipes are the 
same diameter and their axes intersect, one profile may 
be used for the entire operation. 



Note — In the nineteen problems immediately following, the conditions are such that it will be 
necessary to obtain the miter line from the data given by the operation of raking before the straight- 
forward work of laying out the patterns can be begun. However, as certain parts of the work of 
raking the miter line and of laying out the pattern are common to both operations, the two are usually 
carried along together, and therefore such points and spaces should be assumed upon the profiles at the out- 
set as will be required in the final stretchout. 

PROBLEM 55. 
To Obtain the Miter Line and Pattern for a Straight Molding Meeting a Curved Molding of Same Profile. 



3-4 




Fig. 21f2.— -The Miter Line between a Curved and a Straight Molding 
of the Same Profile, 



In Fig. 342, let F G J K represent a piece of 
straight molding joining a curved mold, G H I J, 
the profiles of the straight and curved molding being 
the same. To obtain the miter line or line of joint, 
G J, proceed as follows : Draw the profile in line with 
the straight molding, as shown by C D E, and divide 
into any convenient number of parts. From the divi- 
sions in the profile draw lines parallel to F G in the 
direction of the miter indefinitely, and also in the 
opposite direction, cutting the vertical line C E of the 
profile, as shown by the small figures, which corre- 
spond in number to the divisions on the profile. From 
B, the center from which the curved molding is struck, 
draw the line B A through the molding, as shown. 
Transfer the hights of the various points of the profile 
as obtained on the line C E to the line A B, placing the 
point E at the point o of the intersection of the lower 
line of the curved molding with the line A B, all as 
shown by x o. Then, with B as a center, draw arcs 
from the divisions on the line x o, intersecting lines 
of corresponding numbers drawn from the profile 
parallel to the lines of the straight molding. A line 
traced through these intersections, as shown by G J, 
will be the required miter line, and, as will be seec, is 
not a straight line. To obtain the pattern for the 



Pattern Problems. 



153 



straight molding, draw the line L N at right angles to 
it, upon which place the stretchout of the profile C D 
E, as shown by the small figures. At right angles to 
the stretchout line L N, and through the points in it, 
draw the usual measuring lines. With the T-square 
placed at right angles to K J, bring it successively against 



the points forming the miter line G J, and cut lines of 
corresponding number in the stretchout. Then a line 
traced through these points of intersection will form the 
miter end of the pattern shown by L M N 0. The 
methods employed in obtaining the patterns for the 
curved portions are treated in Problem 127. 



PROBLEM 56. 

A T-Joint Between Pipes of Different Diameters. 



In Fig. 343 it is required to make a joint at right 
angles between thf smaller pipe D F G E and the larger 




fig. S43. — A 1-Joint between Pipes of 
Different Diameters. 

pipe HKLI. For this purpose both a side and an 
end view are necessary. As the two pieces forming the 
T are of different sections this problem really consists 
of two separate operations, but as certain steps can be 



used in both operations the following course will be 
most economical. 

At a convenient distance from the end of the 
smaller pipe in each view draw a section of it. Space- 
these sections into any suitable number of equal parts, 
commencing at corresponding points in each, and Get- 
ting off the same number of spaces, all as shown by A 
B C and A 1 B 1 C. From the 
points in A B C draw lines down- 
ward through the body of the 
large pipe indefinitely. From the 
points in A 1 B 1 C 1 drop point onto 
the profile of the large pipe, as 
shown by the dotted lines. For 
the pattern of the smaller pipe the 
requirements are its profile A B 1 
C and the line F 1 " G 1 , which 
is the outline of the surface against which it miters, 
and therefore its miter line. Therefore, take the 
stretchout of A 1 B 1 C and lay it off at right angles 
opposite the end of the pipe, as shown by V W. Draw 
the measuring lines, as shown. Then, with the T-square 
set parallel to the stretchout line, and brought succes- 
sively against the points between F' and G 1 upon the 
profile of the large pipe, cut corresponding measuring 
lines, as shown. Then a line traced through these 
points, as shown from X to Y, will form the end of the 
pattern. 

For the pattern of the larger pipe the stretchout is 
taken from the profile view F 1 G 1 L' and laid off at 
right angles to the pipe opposite one end, as shown by 
1ST P. A corresponding line, M 0, is drawn opposite 
the other end, and the connecting lines M 1ST and P 
are drawn, thus completing the boundary of the piece 
through which an opening must be cut to meet or 
miter with the end of the smaller pipe. According to 
the rule given in Chapter V, a profile and a miter line 
are necessary. The profile F 1 G 1 L 1 has already been 
stated, but no line has yet been drawn in the elevation 



154 



TJte New Metal Worker Pattern Booh. 



of the larger pipe which shows its connection with the 
smaller pipe. This can only be found by projecting 
lines from the points dropped upon F' G 1 through the 
elevation till they intersect with lines previously drawn 
from the profile A B C, as shown between F and G. 
F G then constitutes the miter line. For economy's 
sake, then, the spaces 1 to 4 previously obtained in the 
profile are duplicated upon the stretchout, as shown, to 
which are added as many more (4 to 10) as are neces- 
sary. As the points 1 to 4 have already been dropped 
upon the miter line in its development it is now only 
necessary to drop them parallel to the stretchout line 



into measuring lines of corresponding number, when a 
line traced through the points of intersection, as shown 
by E S T U, will give the pattern of the opening 
required. 

It may be noticed that the development of the 
miter line F G is not really necessary in this case, as 
the points are really dropped from the profile ABC 
right through the elevation till they intersect the 
measuring lines. This happens in consequence of the 
arm or smaller pipe being at right angles to the larger 
one. Different conditions are shown in Problems 57 
and 58 following. 



PROBLEM 57. 
The Joint Between Two Pipes of Different Diameters Intersecting at Other Than Right Angles. 



Let ABC, Fig. 344, be the size of the smaller 
pipe, and YFZ the size of the larger pipe, and let 
H L M be the angle at which they are to meet. Draw 
an elevation of the pipes, as shown byGKIONMLH, 
placing the profile of the smaller pipe above and in line 
with it, as shown, also placing a profile of the larger 
pipe in line with its elevation, as shown.. In this 
problem the profiles of the moldings or pipes are given, 
but the line representing their junction must be ob- 
tained before going ahead. 

To obtain this miter line, first place a duplicate of 
the profile of the smaller pipe in position above the end 
view of the larger pipe, as shown by A 1 B' C, the cen- 
ters of both being on the same vertical line, C N'„ 
Divide both profiles of the small pipe into the same 
number of spaces, commencing at the same point in each. 
From the points in A B C project lines indefinitely 
through the elevation of the arm, as shown. From the 
points in A 1 B 1 C 1 drop lines on to the profile of the 
large pipe, and from the points there obtained cany lines 
across to the left, producing them until they intersect 
corresponding lines in the elevation. A line traced 
through these several points of intersection gives the 
miter line K L, from which the points in the two 
patterns are to be obtained. For the pattern of the 
small pipe proceed as follows : Opposite the end lay off 
a stretchout, at right angles to it, as shown by E F. 
Through the points in it draw the usual measuring 
lines, as shown. In the developing of the line K L 
the points have already been dropped upon the miter 
line. It therefore only remains to carry them into the 
stretchout, which is done by placing the T-square at 




Fig. SU.—Tlie Joint between Two Pipes of Different Diameters 
Intersecting at Other than Right Angles. 



Pattern Problems. 



155 



right angles with the pipe, and, bringing it successively 
against the points in the miter line K L, cut the corre- 
sponding measuring lines, as shown by the dotted 
lines. A line traced through the points thus obtained 
will give the pattern of the end of the arm, as indi- 
cated. 

For the pattern of the large pipe proceed as fol- 
lows : Opposite one end, and at right angles to it, lay 
off a stretchout line, as shown by PS. In spacing off 
this stretchout it is best to transfer the spaces from 4 
to 4 as they exist, as by so doing measuring lines will 
result which will correspond with points already exist- 
ing in the miter line K L, thereby saving labor as in 



the case of the smaller pipe, and also avoiding con- 
fusion. The other points in the profile are taken at 
convenience, simply for stretchout purposes. Draw a 
corresponding line, P T, opposite the other end, and 
connect P R and T S. In laying off the stretchout 
R S, that number is placed first which represents the 
point at which it is desired the seam shall come. For 
the shape of the opening in the pattern, draw measur- 
ing lines from the points 4, 3, 2, 1, 2, 3, 4, as shown, and 
intersect them by lines dropped from corresponding 
points in the miter line. Through the points thus ob- 
tained trace the line TJ V W X, which will represent 
the shape of the opening required. 



PROBLEM 58. 

The Joint Between an Elliptical Pipe and a Round Pipe of Larger Diameter at Other Than 

Right Angles.— Two Cases. 



In Fig. 345 J K L M is the side elevation of the 
round pipe and E F G H that of the elliptical pipe 
joining the larger pipe at the angle F G J. In the 



Wj 




Fig, S45 —The Joint between an Elliptical Pipe and a Round Pipe of Larger 
Diameter at Other than Right Angles. First Case. The Major Axis of the 
Elliptical Pipe Crossing the Round Pipe. 

end elevation T S I shows the profile of the round 
pipe and UEST the intersection of the elliptical 



pipe whose profile is shown at A B C D and 1ST P Q, 
respectively, in the side and end views. From an in- 
spection of the drawings it will be seen that the side 
elevation shows the narrow view of the elliptical pipe, 
while the end elevation shows its broad view, or in 
other words, that the profile of the elliptical pipe is so 
placed that its major axis crosses the round or larger 
pipe. In Fig. 346 the elevations show 
the same pipes intersecting at the same 
angle, but with the difference that the 
profile of the elliptical pipe is so placed 
that its minor axis crosses the round 
pipe. The reference letters and figures 
are the same in the two drawings and 
the following demonstration will apply 
equally well to either : 

By way of getting ready to lay out 
the miter, it will first be necessary to 
obtain a correct elevation of the miter 
line or intersection between the two 
pipes, as shown, from H to G. To do 
this divide the two profiles ABCD 
and N P Q into the same number of 
equal parts, commencing at the same 
points in each. Draw lines from the 
points in N P Q, parallel with U 
T, cutting T S. In a similar manner draw lines in- 
definitely from the points in A B C D, parallel with 



156 



The New Metal Worker Pattern Booh. 



HE, as shown. From the points in T S draw lines 
parallel with M J, which produce until corresponding 
lines from the two profiles intersect. Through the 
several points of intersection thus obtained draw the 



right angles with H E, or parallel with V W, and 
brought successively against the points in the miter 
line H G, cut corresponding measuring lines. A line 
traced through the points thus obtained, as shown by 



w» 




Fig. $46. — Second Case. The Minor Axis of the Elliptical Pipe Crossing the 

Round Pipe. 



miter line H G. For the pattern of E F Gr H proceed 
as follows : On E F extended, as V W, lay off a 
stretchout of profile ABCD, through the points in 
which draw the usual measuring lines at right angles 
to the stretchout line. With the T-square placed at 



X Y Z, will give the miter cut required, and VWXY 
Z shows the entire pattern. 

The method of obtaining the shape of the open- 
ing in the round pipe is exactly similar to that de- 
scribed in the several preceding problems. 



PROBLEM 59. 



A T-Joint Between Pipes of Different Diameters, the Axis of the Smaller Pipe Passing to One Side of That of the Larger. 



The principle here involved and the method of 
procedure are exactly the same as in Problem 56, but 
the whole of the profiles must be used instead of the 
halves, because the two axes or center lines of the 
pipes do not intersect. 

In Fig. 347, let A B C be the size of the small 
pipe and F 1 H 1 M 1 be the size of the large pipe, between 
which a right-angled joint is to be made, the smaller 



pipe being set to one side of the axis of the large pipe, 
as indicated in the end view. Draw an elevation, as 
shown byDFILMKGE. Place a profile of the 
small pipe above each, as shown by A B C and A' B' 
C, both of which divide into the same number of equal 
parts, commencing at the same point in each. Place 
the T-square parallel to the small pipe, and, bringing 
it successively against the points in the profile A 1 B 1 



Pattern Problems. 



157 



C, drop lines cutting the profile of the large pipe, as 
shown, from F 1 to H' ; and in like manner drop lines 
from the points in the profile ABC, continuing them 
through the elevation of the larger pipe indefinitely. 
For the pattern of the small pipe set off a stretchout 




--10 

-■li 

--12 
_ 13 R 



Fig. 31,7. — A T-Joint between Pipes of Different Diameters, the Axis 
of the Smaller Pipe Passing to One Side of that of the Larger. 



line, V "W, at right angles to and opposite the end of 
the pipe, and draw the measuring lines, as shown. 
These measuring lines are to be numbered to corre- 
spond to the spaces in the profile, but the place of 
. beginning determines the position of the seam in the 



pipe. In the illustration given the seam has been 
located at the shortest part of the pipe, or, in other 
words, at the line corresponding to the point 10 in the 
section. Therefore commence numbering the stretchout 
lines with 10. Place the T-square at right angles to the 
small pipe, and, bringing the blade suc- 
cessively against the points in the pro- 
file of the large pipe from F 1 toH 1 , cut 
the corresponding measuring lines, as 
shown. A line traced through the points 
thus obtained, as shown by X Y Z, will 
form the end of the required pattern. 

For the pattern of the large pipe, 
lay off a stretchout from the profile 
shown in the end view, beginning the 
same at whatever point it is desired to 
locate the seam, which in the present 
instance will be assumed on a line 
corresponding to point 13 in the pro- 
file. After laying off the stretch- 
out, opposite one end of the pipe, 
as shown on R, draw a corresponding line opposite 
the other, as shown by N P, and connect N O and P 
E, thus completing the outline of the pattern, through 
which an opening must be cut to miter with the end 
of the smaller pipe. In spacing the profile of the large 
pipe, the spaces in that portion against which the small 
pipe fits are made to correspond to the points obtained 
by dropping lines from the profile of the small pipe 
upon it, as shown by 1 to 7 inclusive. This is done 
in order to furnish points in the stretchout correspond- 
ing to the lines dropped from the profile A B C, as 
shown. No other measuring lines than those which 
represent the portion of the pipe which the small pipe 
fits against are required in the stretchout. Accord- 
ingly the lines 1 to 7 inclusive are drawn from R, 
as shown, and are cut by corresponding lines dropped 
from ABC. A line traced through the several points 
of intersection gives the shape ST U, which is the 
opening in the large pipe. If it be necessary for any 
purpose to show a correct elevation of the junction be- 
tween two pipes, the miter line F H G is obtained by c 
intersecting the lines dropped from ABC with cor- 
responding lines carried across from the same points 
obtained on the profile F 1 IT, by dropping from A 1 B 1 
C 1 , explained in Problems 56, 57 and 58, and all as 
shown by the dotted lines. 

As remarked in Problem 56, this line is not abso- 
lutely necessary, but is of great advantage in illustrat- 
ing the nature and principles of the work to be done. 



158 



Tlie Neiv Metal Worker Pattern Book. 



PROBLEM 60. 

A Joint at Other Than Right Angles Between Two Pipes of Different Diameters, the Axis of the Smaller Pipe 

Being Placed to One Side of That of the Larger One. 



In Fig. 348, let C B 1 A 1 be the size of the smaller 
pipe, and D' E 1 I' the size of the larger pipe, between 
which a joint is required at an angle represented by W 
F K, the smaller pipe to be placed to 
the side of the larger. Draw an eleva- 
tion of the pipes joined, as shown by 
VDGHIKFW. As in the preced- 
ing problems, the miter line or line giving 
a correct elevation of the junction of 
the pipes must be developed before the 
actual work of laying out the miter pat- 
terns can be begun, therefore place a 
profile or section of the arm in line with 
it, as shown by C B 1 A', and opposite 
and in line with the end of the main 
pipe draw a section of it, as shown by 
D 1 E 1 F. Directly above this section 
draw a second profile of the small pipe, 
as shown by C B A, placing the center 

of it in the required position relative 

to the center of- the profile of the large 
pipe. Divide the two profiles of the 
small pipe into the same number of equal spaces, 
commencing at the same point in each. From the 
divisions in C B 1 A' drop lines parallel to the lines of 
the arm indefinitely. From the divisions in C B A 
drop lines until they cut the profile of the large pipe, 
as shown by the points in the arc D 1 E'. From these 
points carry lines horizontally to the left, producing 
them until they intersect the corresponding lines from 
C B 1 A 1 . A line traced through these points of in- 
tersection, as shown by D E F, will be the miter line 
between the two pipes. For the pattern of the arm 
proceed as follows : Lay off a stretchout at right 
angles to and opposite the end of the arm, as shown 
by B P, and through the points in it draw the usual 
measuring lines. Place the T-square at right angles 
to the arm, and, bringing it successively against the 
points already in the miter line, cut the correspond- 
ing measuring lines. A line traced through these 
points, as shown by UTS, will form the required 
pattern. 

For the pattern of the main pipe draw a stretch- 
out line opposite one end of it, as shown by M 0, 
numbering the divisions in it with reference to locat- 
ing the seam, which can be placed at any point 



desired. The spaces of the profile between D' and E 1 
should be transferred to the stretchout point by point 
as they occur, as by so doing measuring lines will be 




Fig. S4S.—A Joint at other than Right Angles betiveen Two Pipes of 
Different Diameters, the Axis of the Smaller Pipe being Placed to 
One Side of that of the Larger One. 

obtained which will correspond to the points already 
in the miter line. Draw a line corresponding to the 



Pattern Problems. 



159 



stretchout line opposite the other end of the pipe, as 
shown by L N, and connect L M and N 0. Through 
the points in the stretchout line corresponding to the 
points between D' and E 1 of the profile draw measuring 
lines, as shown by 10, 11, 12, 13, 14, 15 and 1. Place 
the T-square at right angles to the main pipe, and, bring- 



ing the blade against the points in D E F successively, 
cut the measuring lines of corresponding number, all 
as shown by the dotted lines. A line traced through 
these points of intersection, as shown near the middle 
of M L 1$ O, will give the shape of the opening to be 
cut in the pattern of the main pipe. 



PROBLEM 6i. 

The Patterns for a Pipe Intersecting: a Four-Piece Elbow Through One of the Miters. 



In Fig. 349, let A B C D E E' D' C B' A 1 repre- 
sent the four-pieced elbow in elevation, F G H J its 




Fig. $49. — Pattern for the Pipe Intersecting an Elbow Through One 
of the Miters. 

profile, and KLMN the elevation of the pipe which 
intersects the elbow through a miter joint. In line with 



the pipe draw the profile of same, as indicated by P 
R S. Extend F H of the profile of elbow, upon which 
as a center line draw another profile of the small pipe, 
as shown by O 1 P 1 R 1 S 1 . Divide both profiles of the 
small pipe into the same number of parts, commencing 
at the same points in each, as S and S 1 . Now parallel 
to F P 1 of the profile draw lines from the points in O 1 
P 1 R 1 S 1 intersecting the profile F G H J, as shown. 

A profile should properly be drawn in its correct 
relation to the part of which it is the section. As the 
part C D D' C is about to be considered first, the pro- 
file should be placed with its center line F H at right 
angles to C D ; but as in a regular elbow of any num- 
ber of pieces the miter lines all bear the same angle 
with the sides of the adjacent pieces, the profile may 
for convenience be placed in proper relation to one of 
the end pieces, after which lines may be carried from 
it parallel to the side it represents to the miter line, 
thence from one miter line to another, always keeping 
parallel to the side, continuing this throughout the 
entire elbow if necessary. Therefore parallel to D E 
of the elevation draw lines from the points in G H of 
the profile, cutting the miter line D' D, and continue 
these lines parallel to D C and C B. From the points 
in the profile OPES draw lines parallel with L K 
intersecting lines of corresponding numbers drawn from 
G H. A line traced through these intersections will 
give the miter line K Z 1ST. From the point Z in 
the miter line carry a line back to the profile of the 
pipe, as indicated by Z a. This gives, upon the pro- 
file of the pipe, the point at which the miter line K 1ST 
crosses the miter line of the elbow C C 1 , so that it can 
be located upon the stretchout line, where it is 
marked a 1 . 

For the pattern of the pipe KLMN proceed as 
follows : At right angles to K L draw the line 
M 1 M 2 , upon which lay off the stretchout of P 
R S, as shown by the small figures, through the 
points in which, and at right angles to it, draw the 



160 



Tlie New Metal Worker Pattern Book. 



usual measuring lines, which, intersect with lines of 
corresponding numbers drawn at right angles to the 
line of the pipe L K from the intersections on the 
miter line KZN. A line traced through the inter- 
sections thus obtained, as shown by M 1 N 1 K 1 W W, 
will be the recpuired patternfor the intersecting pipe. 

To avoid confusion of lines in developing the 
patterns of the intersected pieces of the elbows a 
duplicate of those parts, as shown by B C D D' C B', 
is given in Fig. 350, in which the miter line K Z N is 
also shown. The profiles FGHJ and O 1 P 1 R 1 S' are 
presented merely to show the relationship of parts, as 
the patterns are obtained from the miter line KZN, 
in connection with the stretchout of as much of the 
profile as is covered by the intersection. It is not 
necessary to include in this operation the entire elbow 
pattern, therefore only such a part of the pattern will 
be developed as is contained in profile from V to H. 

For the pattern for that portion of elbow shown 
in elevation by U Z N or V H of profile, proceed as 
follows : At right angles to C D of elevation draw 
the line R S, upon which lay off the stretchout of V H 
of the profile, as indicated by the small figures, through 
which draw the usual measuring lines at right angles 
to it, which intersect with lines of corresponding num- 
bers drawn from the intersections on the miter line 
Z N at right angles to C D. Trace a line through the 
intersections thus obtained, as shown by U 3 Z 3 N 1 . 
Then will U 3 Z 3 represent the pattern for that part of 
elbow shown in elevation by U Z, and Z 3 N 1 be the 
pattern for the cut on the miter line Z N. The pat- 
tern for the other half of opening shown by N 1 X V 2 S 
is simply a duplicate of the half just obtained re- 
versed. Then X N' Z 3 shows the shape to be cut out 
of what would otherwise be a regular elbow pattern. 
The point a in the profiles O 1 S 1 and V II is so near 
the line drawn from the point 4 that separate lines 
are not shown, and on this account when obtaining 
the shape of K 1 Z 1 the points 4 and a are shown on the 
same line. 

In order to show that the pattern is produced by 
the regular method — that is, by the intersection of 
points from the miter line into lines of corresponding- 
number in the stretchout — it should be noted that the 



profile and stretchout of the piece already developed 
is properly designated by the figures 1, 2, 3, a, 5, 6, 
while that of the piece next to be considered is 
properly designated by the figures 1, 2, 3, 4a, 5, 6, the 
point 4 not occurring in the first piece at all, while the 
points 4 and a both fall upon the same line in the stretch- 
out of the second piece, all of which is clearly shown. 
The pattern of the cut on the miter line K Z is 
obtained in the same manner as for Z N. At right 
angles to B C draw the line R 1 S 1 , upon which place 
the stretchout of V H of the profile, as shown. 








N R // 1^ 




B' D 

ELEVATION 



Fig. 350. — Patterns for the Pieces of an Elbow Intersected by the 

Pipe. 

Through the points in the stretchout and at right 
angles to same draw the usual measuring lines, which 
intersect with lines of corresponding numbers drawn 
from the intersections on the miter line K Z, at right 
angles to B C. Trace a line through the intersections 
thus obtained, as shown by K 1 Z 1 IT". Then will 
Z 1 IP represent the pattern for that part of elbow 
shown in elevation by Z U, and Z 1 K' be the pattern 
for the cut on the miter line Z K. The pattern for 
the other half of opening shown by K' X 1 V 1 S 1 can be 
obtained by duplication. Then will X 1 K 1 Z 1 repre- 
sent the shape to be cut out of the regular elbow 
pattern. 



Pattern Problems. 

PROBLEM 62. 

The Pattern for a Gable Molding; Mitering- Agfainst a Molded Pilaster. 



16] 



Let N X V E in Fig. 351 be the elevation of a 
gable molding of which A B C D is the profile, and 
K M L be the elevation of a molded pilaster against 



straight from N to E, as would be the case if the side 
of the pilaster were perfectly flat and projected further 
than the gable molding. It will therefore be neces- 




Fig. 251. — The Pattern for a Oable Molding Mitering Against a Molded Pilaster. 



which it is required to miter. The profile of the 
pilaster is shown by J I H in the plan, where a profile 
of the gable mold A 1 B 1 C D 1 is also shown and so 
placed as to show the comparative projection of the 
various points in each. By an inspection of the plan 
and elevation it will be seen that the miter line or joint 
between the molding and the pilaster will not be 



sary to first obtain a correct elevation of the miter, 
after which the pattern can be obtained in the usual 
simple manner. 

To do this divide the profiles in the plan and 
elevation into the same number of equal parts, com- 
mencing at the same points in each, as shown by the 
corresponding figures. From the divisions in the pro- 



162 



The New Metal Worker Pattern Booh. 



file in plan carry lines to the left, parallel to H A ! , 
until they cut the side of the pilaster H I J, as shown. 
From these intersections drop lines at right angles to 
H A 1 indefinitely, as shown. From the divisions in 
the profile in elevation draw lines parallel to N X 
until they intersect corresponding lines drawn from 
H I in plan. A line traced through these intersections, 
as shown by N F E, will be the required miter line, or 
intersection of the gable molding with the upright 
pilaster at the angle O 1ST X. 

Foi the pattern of the gable molding proceed as 
follows : At right angles to the lines of the gable 
molding draw the stretchout line A 2 D 2 , upon which 
place the stretchout of the profile of the gable mold- 
ing, through which draw the usual measuring lines, 
which intersect with lines of corresponding number 
drawn from the points in the miter line N FE at right 



angles to the lines of the gable molding. A line traced 
through these intersections, as shown by N G E Q , will 
be the required pattern. 

Although the roof strip A B of the gable mold- 
ing is perfectly straight, points will have to be intro- 
duced between A and B for the purpose of ascertaining 
the shape of the cut from X to F, its intersection with 
the side of pilaster. The simplest method of obtain- 
ing these points is to derive them from the points be- 
tween B 1 and C, as shown by 0' to 5' in the plan. 
They can then be transferred to their proper place in the 
stretchout, as shown, between A = and B\ By so doing 
points of like number fall in the same place on the pro- 
file H I, and the vertical lines dropped therefrom can be 
intersected with F N for the pattern of the roof strip 
and with the other lines from F to E for the pattern of 
the face of the mold, all of which is clearly shown. 



PROBLEM 63. 

The Patterns for an Anvil. 



It frequently occurs that sheet metal reproductions 
' of various emblems or tools are desired for use as orna- 
ments or signs. In the following problem is shown 
how the various pieces necessary to form an anvil may 
be obtained. The description, of course, only applies 
to the several sides, as a representation of the horn 
can only be obtained by hammering or otherwise stretch- 
ing the metal. 

In Fig. 352 is shown a side and end elevation and 
two plans of the anvil, exclusive of the horn, the plans 
being duplicates and so placed as to correspond re- 
spectively with the side and the end views. Before 
the pattern of the side piece JNBG can be developed 
the line P Q E S, which is the result of the miter- 
ing together of the two forms shown by U V W of the 
plan and Z X T of the end view, must be obtained. 
Therefore divide the curved portion Z X T of the 
profile of the side into any convenient number of 
equal spaces, as shown by the small figures, and from 
the points thus obtained drop lines vertically cutting 
W V U', the profile of the gore piece. Transfer the 
points thus obtained on W V U 1 to¥YU of the 
other plan, and from these points erect lines vertically 
through the elevation of the side, and finally intersect 
them with lines of corresponding number drawn from 
the points originally assumed in Z T. Then a line 
traced through the points of intersection, as shown by 
O P Q E S, will be the required miter line. 



To obtain the pattern of the side, first lay off a 
stretchout of the profile Z T, as shown, upon v q, through 
the points in which draw the usual measuring lines. 



M 






M 1 










/1 

M 




1 


2 

3 


i i \ 


1 1 1 ELEVA 


TION 


i i 


1 '! 




i i 


1 '! 


I 

n 


: v' 


i i i 

i i i 


1 

1 ' 
1 1 1 

1 ! 


m 


m' 


i i i 
i i i 

1 1 


1 1 ! 
1 1 1 

1 ijL 




-1 


1 1 1 


1 / 

1/ PATTER 


N FOR 


■\ 1 
\i 1 


/ 




/, FRONT 

Y 


r° Xi 


<r 





D 



J 1 




Fig. 353. — Patterns for the End Pieces of an Anvil. 

With the T-square placed parallel with J Gr drop lines 
from the points in the profile Z T cutting the outlines 



Pattern Problems. 



163 



of the side from J to a and G to b, and also cutting the 
miter lines of the gore piece Q S (which last opera- 
tion has really been done in the raking operation above 
described). Placing the T-square parallel with v q, 
bring it successively against the points in the several 
miter lines of the side elevation and cut corresponding 
measuring lines ; then lines traced through the points 



the raking operation, lay off a stretchout of the same 
upon any line running at right angles to the form of 
this piece, as shown upon IT 2 W\ As the points have 
already been dropped from the profile to the miter lines 
in the operation of obtaining them, it only remains to 
place the i -square parallel to II 2 W 2 and bring it suc- 
cessively against the points in Q and Q S, cutting 




F' C 

Fig. 352. — Patterns for the Side, Gore Piece and Bottom of an Anvil. 



of intersection, as shown, from j to c, o to X, X to s and 
g to d, will give the pattern for the lower portion of the 
side. As that part of the side from "Y to Z of the pro- 
file is straight and vertical, that portion of the pattern 
shown on the stretchout line from X to q can be made 
an exact duplicate of that part of the elevation shown 
by a M X B E b, all as shown. 

For the pattern of the gore piece, U V W is the 
profile and Q and Q S are the miter lines. By means 
of the points previously obtained upon the profile in 



corresponding measuring lines ; then lines traced 
through the points of intersection, as shown by IP 
Q' and Q 1 "W 2 , will give the pattern for the gore 
piece. 

For the end pieces of the anvil, NMaKJ and 
B E b H G of the side elevation become the profiles, 
and Z T and Z' T" are the miter lines. Therefore, to 
obtain the pattern of either of these pieces, inde- 
pendently of the preceding operations, space the curved 
portion of its profile into any convenient number of 



164 



Tlie New Metal Worker Pattern Booh. 



spaces, and lay off a stretchout of the same upon any 
line at right angles to T T". Carry lines from the 
profiles parallel to N B, cutting the miter lines, thence 
at right angles to T T 1 , cutting corresponding measur- 
ing lines. To avoid confusion of lines the operation 
of obtaining the patterns of the end pieces has been 
shown separately in Fig. 353, in which J N N" 1 J 1 is 
an elevation of the front end and G B B 1 G' that of 
the back end. The points made use of, however, 
upon their profiles, in Fig. 352, are such as were ob- 



tained there in cutting the pattern of the side ; therefore 
their stretchouts must be transferred point by point to 
the stretchout lines ; E D of Fig. 353 being the stretch- 
out of N a J in Fig. 352 and B A of Fig. 353 being 
that of B b G. In consequence of the above the points 
upon the miter line Z T are such as were originally 
obtained there by spacing, and have been transferred 
to the lines N J, W J 1 , B G and B 1 G". The remainder 
of the work is shown sufficiently clear to need no 
further explanation. 



PROBLEM 64. 

The Pattern for a Gable Cornice Mitering Upon an Inclined Roof. 



In Fig. 35-i let A B C G represent one side of 
the gable molding and NOM its profile. H B F E 
represents the horizontal molding, and D C the upper 
line of roof. The profile of this molding is shown by 
K L, and the inclined roof by K J. Before the pat- 
tern for gable can be described it will be necessary to 
obtain an elevation of the intersection of the gable 
cornice with the inclined roof between C and B to be 
used as the miter line. 

The first step to be taken in obtaining this miter 
line is to draw the profile of gable cornice, P Q B, 
directly over and in line with the profile of 
the horizontal molding J K L, as shown. 
Divide both profiles M and P E, into the 
same number of parts. From the points in 
M draw lines parallel with the rake, ex- 
tending them indefinitely in the direction of 
C B. From the points in P R drop lines 
upon the roof line J K, and from the points 
of intersection in J K carry lines horizontally 
across to the elevation, intersecting" them with 
lines of corresponding number previously 
drawn from M. Through the points of 
intersection trace a line, which will be at once 
the correct elevation of the miter and the 
miter line from which to obtain the pattern. 
If the pattern of gable at A G is required, in 
connection with that at the foot, extend the 
lines from points in M O to the miter line A G. 

To obtain the pattern of A B C G, pro- 
ceed as follows : At right angles to A B of 
gable lay out a stretchout of M 0, as shown by S T, 
through the points in which draw the usual measuring 
lines. Place the T-square at right angles to the gable 
line A B, and, bringing it successively against the 



several points in A G and C B, cut corresponding 
measuring lines, all as indicated by the dotted lines. 

V 




ELEVATION 



Fig. S54.- 



-Method of Obtaining Miter Line and Pattern for a Gable Cornice 
Mitering Upon an Inclined Roof. 

Thus the line U X of pattern is of the same length as 
A B of elevation, and V W of pattern the same length 
as G C of elevation, etc. It is evident that the various 
lines in pattern are of the same length as lines of corre- 



Pattern Problems. 



165 



sponding number in elevation. Through the points 
obtained in the pattern trace lines as indicated by UV 



and W X. Then U V W X is the pattern for the 
part of gable shown by A B C G in elevation. 



PROBLEM 65. 

The Pattern for the Molding on the Side of a Dormer Mitering Against the Octagonal 

Side of a Tower Roof. 




Fig. S55.—The 
Mitering 



Pattern for the Molding on the Side of a Dormer 
Against the Octagonal Side of a Tower Roof. 



Let F J H G in Fig. 355 represent a half elevation 
of a portion of the tower roof corresponding to A B C 
D E of the plan ; also let U O P B S be the side eleva- 
tion of a dormer cornice for which the pattern is re- 
quired, KLIN being half the front elevation of the 
dormer, and profile of the molding. The first step 
before the pattern can be described is to obtain a cor- 
rect elevation of the miter line or intersection of the 
cornice with the oblique side of the tower, as shown by 
UTS. To obtain this miter line proceed as follows : 
On A E of the plan extended as a center line, as ISP K 1 , 
draw a duplicate of the half front elevation correspond- 
ing to the half E D of the plan as shown by K' L" M' 
N 1 . Now divide the two profiles of the return mold- 
ing into the same number of parts, commencing at the 
same points in each, as shown by the small figures. 

From each of the points in the upper front eleva- 
tion carry lines parallel with 17 K cutting the side 
F J of the tower, and for convenience in obtaining the 
miter line extending them into the figure, as shown. 
From the intersections obtained on the side of the tower, 
as shown by the small figures in IT S, drop lines par- 
allel to the center line F G until they cut the miter 
line A D of the plan. 

From the points in A D draw lines parallel with 
C D (the oblique side), extending them indefinitely to- 
ward A C. Now from the points in the lower front ele- 
vation K 1 L 1 M 1 draw lines parallel with A K 1 , producing 
them until they meet or intersect lines of correspond-, 
ing numbers, just described. A line traced through 
these intersections, as shown by IT 1 IP T 1 S', will give 
the shape of the miter line as it will appear in plan, IP 
T 1 S 1 showing that portion of the intersection which 
occurs upon the oblique side of the tower roof. 

From the points of intersection in the miter line 
IF T 1 S 1 of the plan erect lines parallel to A B, 
producing them until they intersect lines of correspond- 
ing numbers drawn from the profile K L M N through 
the side elevation. 

A line traced through these intersections, as shown 



166 



TJte New Metal Worker Pattern Book. 



in U T S of the elevation, will be the miter line in eleva- 
tion, formed by the junction of the return with the 
oblique side of the tower A D C of the plan. 

To obtain the pattern proceed as follows : At 
right angles to U of the elevation draw the line V 
W, as shown, upon which lay off the stretchout of K 
L M of the front elevation, as shown by the small 



figures, through which draw the usual measuring lines, 
which intersect with lines of corresponding numbers 
drawn at right angles to U of the elevation from the 
points of intersections in the miter line UTS and from 
the points of intersections in the profile OPE. A 
line traced through these intersections, as shown by IP 
T S 2 R 2 P 2 2 , will be the required pattern. 



PROBLEM 66. 

The Pattern for an Inclined Molding Mitering Upon a Wash Including' a Return. 



As a feature of design, it frequently occurs that 
a belt course between stories is carried around pilasters 
which occur between all the windows of a front, and 



between the pilasters and partly upon the wash of the 
returns at the sides of pilasters. Such a condition of 
affairs presents some interesting features and is shown 




PLAN 



Fig. 356.— The Pattern for an Inclined Holding Mitering Upon a Wash Including a Return. 



which rise from the foundations to the main cornice. 
In a certain instance, small gables or pediments were 
introduced between the pilasters in such a manner that 
the miters at the foot of the gables came partly upon 
the roof or wash of that part of the belt course lying 



in Pis. 356, of which A B C is the front elevation. 
D E Gr shows the plan of the belt course, upon which 
the foot of the gable mold is required to miter in the 
vicinity of F. The gable mold, of which J C is the 
elevation and H the profile, is required to meet the 



Pattern Problems. 



167 



level cornice at the angle C J M, its top line starting 
from the point J. In this instance, as in many others, 
the first requisite is that of obtaining a correct eleva- 
tion of the miter between the gable mold and the three 
washes. To facilitate this operation it will be neces- 
sary to draw a side view which will show the compara- 
tive projection of the gable mold, the pilaster and the 
belt cornice from the face of the wall, as shown at the 
right. Divide both profiles of the gable mold into the 
same number of spaces, as shown by the small figures. 
From the points in the profile II of the elevation carry 
lines parallel to C J, extending them across the line 
J L indefinitely. From the points in IT drop lines 
cutting the profile of the wash of the belt course P 
so far as they fall within its projection. From the 
points in P carry lines horizontally across the eleva- 
tion till they intersect lines of corresponding number 
previously drawn from the profile H. Inspection will 
show that only the lower portion of the profile will 
miter upon the main wash and that, therefore, the 
above operation can be begun with advantage at point 
14 and continued until a line traced through the points 
of intersection crosses the line J K, which is really 
the profile of the wash of the return. This, as will be 
seen, occurs at S, which point can be carried back to 
profile H (shown at x), where it will be subsequently 
needed in obtaining the stretchout of the gable mold. 
Above the point x of the profile all points will fall 
against the wash of the return J K until the projec- 
tion of the mold carries them across the forward miter 
of the return (J 1 K 1 in plan), after which they will fall 
upon the wash in front of the pilaster. This point of 
crossing can be found by reversing the operation above 



described, thus: From points upon J K above S 
carry lines horizontally across to side view, intersecting 
them with lines of corresponding number dropped 
from profile H 1 until a line traced through points of 
intersection, shown by S 1 T', crosses the line O 1 P 1 , as 
shown at point T', which point happens to coincide 
with point 4 of profile. From point 4 of profile H' 
lines are dropped upon 0' P 1 , from which they are 
carried horizontally across as in the first part of 
the operation till they intersect with lines of cor- 
responding number drawn from points in profile II, 
as shown from T to N. Then the line J 1ST T S L 
will be a correct elevation of the required intersection 
and can be used as the miter line from which to obtain 
the final pattern of the gable mold. With this as a 
miter line and H as a profile the remaining operation 
is performed in the usual manner. Upon any line, as 
V W, drawn at right angles to J C lay off a stretchout of 
the profile of gable mold. In obtaining this stretchout 
the position of point x must be obtained from profile 
H, while from profile II 1 is obtained the position of 
point Q, which shows the point at which the roof 
piece of the gable mold passes beyond the side of 
the pilaster, shown best at J' in the plan. With the 
"["-square placed parallel to V AY, and brought success- 
ively against the points in J N T S L, cut measuring 
lines of corresponding number. A line traced through 
the points of intersection, as shown by J' N 1 S 1 L 1 , will 
give the required pattern. The plan view of the in- 
tersection is shown at ISP L", with some of the lines of 
projection used in obtaining it, merely to assist the 
student in seeing the relation of parts, but is not- 
necessary in the actual work of obtaining the pattern. 



PROBLEM 67. 



The Pattern for a Level Molding: Mitering Obliquely Against Another Level Molding - of Different Profile. 



In Fig. 357 is shown the plan and a portion of 
the side view of a bay window. In the side view is 
also shown the section of a lintel molding, shown 
indefinitely by C D E F of the plan, which it is re- 
quired to miter against the oblique side of the large 
cove under the bay window indicated by B C F of the 
plan. In Fig. 358 is shown an enlarged plan of the 
particular portion in which the miter occurs, the angle 



BCD being the same as B C D of Fig. 357. In Fig. 
35S, A B F G represents the base of the window and G E 
D C the lintel cornice. The profiles are shown respect- 
ively at Y and X. The lintel molding is continued in 
the direction of F G until it intersects the base of the 
window between G and C. 

In order to obtain the pattern of that part of the 
lintel molding which abuts against the base of the 



16S 



The Neio Metal Worker Pattern Booh. 



window indicated from G to C, it is first necessary to 
obtain the plan of the intersection or shape of the 



line with the profile Y draw a duplicate of X, as shown 
at Z. In placing the profile Z in position it must be 



N, M 




Fig. S57. — Plan and Sectional View 
of a Level Molding Mitering 
Obliquely Against Another Level 
Molding of Different Profile. 



Fig. S58. — Method of Obtaining the Pattern of the Lintel Molding Shoivn in Fig. 35?. 



miter line, as shown in plan by G II C F. To obtain 
this miter line proceed as follows : Opposite to and in 



remembered that as nights are all to be compared the 
vertical lines of each profile must be placed parallel 



Pattern Problems. 



169 



and their upper ends turned in the same direction. 
Therefore, the back or line 1 13 of the profile Z is 
placed parallel to B C, which represents a vertical line 
with reference to the profile Y, and the point 12 is 
placed exactly opposite the point J, according to the 
requirements of the side view, Fig. 357. Divide the 
profiles X and Z into the same number of parts, as in- 
dicated by the small figures in each. From the points 
thus obtained in profile Z carry lines at right angles to 
B C, cutting K J of profile Y, as shown. With the 
T-square placed parallel with the line B C of the plan 
of the window carry lines from the points on the pro- 
file K J in the direction of G and F; also draw lines 
from the points in the profile X parallel to G E, cutting 
the lines of corresponding number drawn from the pro- 
file Y. A line traced through points of intersection, 
as shown by G H C F, will give the miter line, as 
shown in the plan. 

While the curved portions of the jirofiles X and 
Z have been divided into such a number of spaces as 
will answer the purpose of an ordinary miter, it will 
be noticed that the plane surfaces between points 2 and 
3 and 11 and 12 intercept so much of the curve of K 



J as to produce a curve between those points in the 
pattern. Therefore, for accuracy it is necessary to 
subdivide those spaces on K J, as indicated by a b and 
c d e there shown. These points must be dropped back 
to the profile Z, and the spaces thus produced transferred 
to the stretchout line L M, all as indicated. The lines 
indicated by the small letters in K J have only been 
drawn part way in the engraving to avoid confusion, 
and the measuring lines produced by these points in 
the stretchout have been shown dotted for the sake of 
distinction. These points are then intersected with the 
surfaces to which they belong in the miter line G H 
C, as shown between 2 and 3 and 11 and 12. 

For the pattern of the lintel molding first draw a 
line at right angles to it, as shown by L M, on which 
line lay off a stretchout of the profile X, as indicated 
by the small figures. Through the points thus ob- 
tained draw the usual measuring lines. With the 
T-square placed parallel with the stz-etchout line L M 
carry lines from the points in G H C F to measuring 
lines of corresponding number, when a line drawn 
through the points of intersection, as shown by NOP, 
will complete the pattern. 



PROBLEM 68. 



The Patterns for a Square Shaft of Curved Profile Mitering- Over the Peak of a Gable Coping; Having- 

a Double Wash. 



Let A B C in Fig 359 be the front elevation and 
D E F G H be the side elevation of a coping to sur- 
mount a gable, the profile of the top of which is shown 
at D K II in the side elevation. Also let M X P 
be the elevation of a square shaft or base, as of a finial, 
having a curved profile, as shown, which it is required 
to miter down upon the top of the cojiing. As the 
matter of drawing the elevations of the shaft in correct 
position upon the washes of the coping is attended with 
some difficulty, the method of obtaining these will be 
briefly described first : Through the lowermost point 
on either side of the front elevation, as P, draw a line 
at the correct angle of the pitch of the coping or gable, 
as shown by Y W, and extend the same to the right 
far enough to permit a section of the coping to be con- 
structed upon it. Upon this line set off the distance 
X W, equal to P L (half the width of the shaft at its 



base), and through the point W draw a line perpendicu- 
lar to Y W ; next through the point X draw a line, 
making the same angle with Y W that D K, of the 
given profile of the coping, does with the horizontal 
line D L 1 , and extend this line to the right till it meets 
the line from W at K', and to the left, making D" K' 
equal to D K. This gives one-half the profile of the 
wash, which is all that is necessary in obtaining the 
elevation. Now from the points in the profile D 1 K 1 
project lines parallel to Y W till they meet the center 
line of the front elevation, and duplicate them on the 
other side of the center line, which will complete the 
front elevation of the gable. From the points in the 
profile D K II erect vertical lines indefinitely, which 
may be intersected with lines projected horizontally 
from the points on center line I L to complete the side 
elevation. Thus a line from point B intersected with 



170 



Tlie New Metal Worker Pattern Booh. 



line from K will give the apex of coping, and a line 
from point Z intersected with lines from D and H will 
give the points E and G, front and back of the washes 
at the apex. 

As the shaft is exactly square, the side elevation 
of it, W W O 1 P', is in all respects the same as that of 



point U ; while the crossing of the side P with the 
top line of coping B C (marked 4) is projected upon 
the center line of the side view, thus giving the point 
Y. This completes the elevations with the exception 
of the lines MUP and M 1 Y P 1 . If the profile of the 
shaft P were a straight line, either slanting or vertical, 




FRONT ELEVATION 



Fig. 359.— The Patterns for a Square Shaft of Curved Profile Mitering Over the Peak of a Gable. 



the front, and is drawn in line with it, as shown by the 
horizontal lines of projection connecting them. This 
having been completed, the point at which its side ' 
M' N' crosses the line E E (marked 4J-) is projected 
back to the front elevation, as shown, thus locating the 



the lines U P and M l Y would be straight lines, because 
they would represent the intersection of two plain sur- 
faces ; but as the profile is a curved line it will be ■ 
necessary to obtain correct elevations of these two lines, 
. as they are essential in obtaining the patterns to follow. 



Pattern Proble 



ms. 



171 



The shaft being square, the miters at its angles are 
plain square miters and are developed by the ordinary 
method, as explained in several problems in the earlier 
part of this chapter. The peculiarity of this problem, 
then, consists in obtaining the miter lines U P and M 1 Y 
and the part of the pattern corresponding to the same, 
which can all be done at one operation, as follows : 

Divide the profiles P and M 1 N 1 into the same 
number of equal parts, and place the stretchout of the 
same upon the center lines extended, as shown at I J 
and I 1 J, 1 through which draw the usual measuring lines 
for subsequent use. From the points in M 1 N' from 
4-J- down drop lines vertically upon the profile of 
coping D K, as shown by the dotted lines, and trans- 
fer the points thus obtained to the profile D' K', from 
which points draw lines parallel to V W, as shown by 
the dotted lines. Intersect these with lines of corre- 
sponding number (2, 3 and 4) drawn horizontally from 
either profile, as shown at 2', 3' and 4', thus obtaining 
the miter line U" P. After the points 1 to 11 of 
the profile have been dropped into the measuring lines 
of corresponding number of the stretchout the points 2', 
3', 4' and 4^-' are also dropped into the measuring lines 
of corresponding number, thus giving the cut U 1 T at 
the bottom of the pattern, which can be duplicated on 
the other side of the center line, thus completing the 
pattern Q E S T U 1 of the front of the shaft. 



From the points 1 to 4 of the profile P draw 
lines parallel to B C cutting the profile D 1 K', as shown 
by the solid lines, and transfer the same to the profile 
of coping D K, and from these points erect perpendicu- 
lar lines (also shown solid) indefinitely, as shown, which 
intersect with lines drawn horizontally from points of 
corresponding number in cither profile, as shown at 2/' 
3" and 4". This will give the correct miter line M 1 
Y. The miters at the sides of piece M 1 W O 1 P" 
are of course the same as those of the front piece, 
therefore after they have been obtained the points 2" 
and 3" are dropped into measuring lines 2 and 3 of the 
stretchout P J 1 , which when duplicated on the other 
side of the center line complete the line Q' Y 1 T', which 
is the bottom cut of the side piece. 

It has been remarked that in obtaining the inter- 
sections between U and P and M 1 and Y horizontal 
lines may be drawn from points in either profile. The 
reason for this is simply that the two profiles P and 
1ST M' are identical and have been divided into the 
same number of equal parts. If a case should occur in 
which the side and face should be dissimilar it must be 
borne in mind that N 1 M 1 is the profile of the face piece 
and its points must be used in obtaining the intersec- 
tions between U and P, while P is the profile of the 
side piece, and its points must be used in obtaining the 
intersections between M 1 and Y. 



PROBLEM 69. 



The Patterns of a Cylinder Mitering with the Peak of a Gable Coping; Having- a Double Wash. 



Let ABCin Fig. 360 be the elevation of a coping 
to surmount a gable, the jDrofile of which is D E F E 1 
D 1 , which, as will be seen, shows a double wash, E F 
and F E 1 . Let M P N be the elevation of a pipe or 
shaft which is required to miter over this double wash 
at the peak of the gable. Before any patterns can be 
developed it will be necessary to first obtain a correct 
elevation of the miter line or intersection of the shaft 
with the coping. To accomplish this proceed as fol- 
lows : In line with the pipe or shaft construct a profile 
of the same, as shown by G 1 L 1 K 1 H 1 , which divide 
into any convenient number of equal parts, and from 
the points thus obtained drop lines vertically through 
the elevation. Draw a corresponding profile, as shown 



by H G L K, directly over the profile of the coping, 
all as shown, which divide into the same number of 
equal parts, beginning to number at a corresponding 
place in the profile, and from the points in it drop lines 
on to the profile of the coping, cutting the washes E F 
and F E', and thence carry the lines parallel to the lines 
of the coping, producing them until they intersect the 
lines dropped from the profile G 1 L 1 K 1 H 1 . Through 
the points of intersection thus obtained trace a line, as 
shown from to P, then OB'P will be the miter line 
in elevation. 

As both halves of the shaft are alike (dividing on - 
the line H L in one profile, and on H 1 L 1 in the other), 
it is really only necessary to use one-half of the profile, 



172 



The New Metal Worker Pattern Book. 



as to use both halves as in the diagram requires the 
additional work of carrying the points from E FE' to 
the center line B B 1 for one-half and then down the 
other side of the gable for the other side. For the 
pattern of the shaft proceed as follows : In line with 
trie end M N of the shaft, and at right angles to it, lay 
off a stretchout of the profile G 1 H 1 K 1 L 1 , as shown by 
R S, in the usual manner, through the points in which 
lines. Commence numbering these 



angles to the lines of one side of the coping, as A B, 
lay off a stretchout of the -wash of the coping E F E 1 , 
all as shown by E 2 F 1 El In this stretchout line set 
off points corresponding to the points in E F E', ob- 
tained by the lines previously dropped from the profile 
G H K L. Place the T-square at right angles to A B, 
and, bringing it against the points in the miter line O 
B 1 , cut lines of corresponding numbers drawn through 
the stretchout E 2 E 3 , all as indicated by the dotted 




Fig S60.—The Patterns of a Cylinder Mitering with the Peak of a Gable Coping Having a Double Wash. 



measuring lines with the figure corresponding to the 
point at which the seam is desired to be, in this case 1. 
Place the J square at right angles to the shaft, and, 
bringing it against the points in the miter line OB'P 
cut the corresponding measuring lines. Then a line 
traced through these points of intersection, as shown 
by T U V W X, will be the miter required. In case 
it should be desired to miter the coping against the base 
of the shaft, the pattern for it may be obtained from 
the same lines in the following manner: At right 



lines. Then a line traced through these points of in- 
tersection, as shown by Z I Y, will be the pattern of 
the wash for the side of the gable A B required to 
miter against the base of the shaft. 

In case the design should call for a shaft octagonal 
in shape, the same general rules would apply. Less 
divisions, however, will be required in the profile, it 
only being necessary to drop points from each of the 
angles of the octagon, as in the case of Problem 33, 
previously given. 



Pattern Problems. 

PROBLEM 70. 

A Butt Miter of a Molding: Inclined in Elevation Against a Plain Surface Oblique in Plan. 



173 




Fig. 361.— A Butt Miter of a Molding Inclined in Elevation Against 
a Plain Surface Oblique in Plan. 



Let A B in Fig. 361 be the profile of a given 
cornice, and let E D represent the rake or incline of 
the cornice as seen in elevation. Let Gr H represent 
the angle of the intersecting surface in plan. The 
first step in developing the pattern will be to obtain 
the miter line in the elevation, as shown by E F. For 
this purpose draw the profile A B in connection with 
the raking cornice, which space in the usual manner, 
as indicated by the small figures. Draw a duplicate 
of this profile, as shown by A 1 B', placing it in 
proper position with reference to the lines of the plan. 
Space the profile A 1 B 1 into the same number of parts 
as A B, and through the points thus obtained carry 
lines parallel to the lines of the cornice, as seen in 
plan, cutting the line G H, as shown. In like man- 
ner draw lines through the points in A B, carrying 
them parallel to the lines of the raking cornice in the 
direction of E F indefinitely, as shown. Place the 
T-square at right angles to the lines of the cornice, as 
shown in plan, and, bringing it against the points of 
intersection in the line G H, carry lines vertically, 
cutting corresponding lines in the inclined cornice 
drawn from the profile A B. Through the points of 
intersection thus obtained trace a line, as shown from 
E to F. Then E F will be the miter line in elevation, 
formed by an inclined cornice of the profile A B meet- 
ing a surface in the angle shown by G IT in the plan. 

At right angles to the raking cornice lay off a 
stretchout of A B upon any line, as K L, and through 
the points draw the usual measuring lines, all as shown. 
Place the T-square at right angles to the lines of the 
raking cornice, and, bringing it against the several 
points in the miter line E F, cut corresponding meas- 
uring lines drawn through the stretchout KL. A line 
traced through these points of intersection, as shown ■ 
from M to N, will be the pattern required. 



PROBLEM 71. 



Patterns for the Moldings and Roof Pieces in the Gables of a Square Pinnacle. 



Fig. 



362 shows the elevation of one of four sim- 
ilar gables occurring in a square pinnacle. The profile 
of the molding is shown at P. The first step is to 
obtain the miter line or elevation of the miter 



shown at K, from which to derive the pattern. Draw 
the profile P in the moldingr, as shown, placing it so 
that its members will correspond with the lines of the 
molding. Draw a second profile, P', in the side view 



174 



The New Metal Worker Pattern Booh. 



of the gable, placing it, as shown in the engraving, 
so that its members will coincide with the lines of the 
side view. Space both of these profiles into "the same 
number of parts in the usual manner, and through 
the points thus obtained draw lines parallel to each of 
the moldings respectively, as shown, until they inter- 
sect, and trace a line through the points of intersec- 
tion, as shown at K. Then K is the line in elevation 
upon which the moldings will miter. Draw the center 
line O M, which represents the miter at the top of the 
gable. 

For the pattern of the molding lay off a stretch- 
out of the profile upon any line, as G H, drawn at 
right angles to the line of the gable in elevation, as 
shown by the small figures. Through these points 
draw measuring lines, as shown. Place the T-square 
parallel to the stretchout line, or, what is the same, at 
right angles to the line of the gable, and, bringing it 
successively against the several points in the miter lines 
O M and K, cut the corresponding measuring lines, 
as shown, and trace lines through the intersections. 
This completes the pattern of the molding, to which 
the piece forming the roof may be added as follows : 
Make L D 1 equal to E D of the side view of the 
gable and set it off at right angles to L B'. In like 
manner, at right angles to the same line, set off A' B 1 



equal to A B of the side view, and draw the lines 
D' A' and A 1 F, as shown. 




Fig. S6S. — Patterns for the Moldings and Roof . 
of a Square Pinnacle. 



PROBLEM 72. 



Pattern for the Moldings and Roof Pieces in the Gables of an Octagon Pinnacle. 



Fig. 363 shows a partial elevation and a portion 
of the plan of an octagon pinnacle having equal gables 
on all sides. The first step in developing the patterns 
is to obtain a miter line at the foot of the gable, as 
shown by L. To do this proceed as follows : Draw 
the profile K, as shown, placing it so that it shall cor- 
respond in all its parts with the lines of the molding 
in elevation. Divide into spaces and number in the 
usual manner, and through the points draw lines par- 
allel to the lines ol the gable toward L, as shown. 
Draw a duplicate profile in the plan, K 1 , so placed as 
to correspond with the lines of the molding in plan. 
Divide it into the same number of spaces, and through 
the points in it draw lines parallel to the lines of the 



plan, cutting the line D F, representing the plan of the 
miter. From the points in D F thus obtained carry 
lines vertically, intersecting corresponding lines drawn 
from the profile in the elevation. A line traced through 
the several points of intersection, as shown by L, will 
be the line of miter in elevation between the moldings 
of the adjacent gables. The center line 1ST forms the 
miter line for the top of the gable. 

For the pattern proceed as follows : Upon any 
line, as E E, drawn at right angles to the lines of the 
gable, lay off a stretchout of the profile, as shown by 
the small figures. Through the points of the stretch- 
out draw the usual measuring lines. Place the T- 
square at right angles to the lines of the gable, and, 



'Pattern Problems. 



175 



bringing the blade successively against the points in 
the two miter lines above described, cut the corre- 



be added by setting off A 1 B 1 at right angles to A 1 C 1 , 
equal in length to A B of the side view. In like man- 




Fig. 363,— Pattern for the Moldings and Roof Pieces in the Gables of an Octagon Pinnacle. 



sponding measuring lines, as shown. Lines traced 
through the points of intersection thus obtained will 
give the pattern of the molding. The roof piece may 



ner, upon the line from M set off D 1 C equal to C D 
of the side view. Then draw F 1 D' B', thus completing 
the pattern. 



176 



Tlie New Metal Worker Pattern Book. 



PROBLEM 73. 

The Pattern for the Miter Between the Moldings of Adjacent Gables Upon a Square Shaft. Formed by 

Means of a Ball. 



In Fig. 364, let A C be one of the gables in pro- 
file and B D the other in elevation, the moldinsrs form- 
ing a joint against a ball, the center of which is at E. 
The first operation necessary will be that of obtaining 
the miter line, or, in other words, the appearance in 
elevation of the intersection of the molding with the 
ball. Place the profile of the mold in each gable, as 
shown at F and II. Divide each of these profiles into 
the same number of equal parts, as indicated by the 
small figures. From the points thus obtained ill F 
drop lines vertically, meeting the profile of the ball, 
as shown from C to J. From the center E of the ball 
erect a vertical line, as shown by E J. From the 
points in C J already obtained carry lines horizontally, 
cutting E J, as shown, and thence continue them, by 
arcs struck from E as center, until they meet lines of 
corresponding number dropped from points in the pro- 
file H parallel to the gable in elevation. Through the 
intersections thus obtained trace a line, as indicated by 
D Gr M. Then DGM will be the miter line in eleva- 
tion. To develop the pattern for the molding, first lay 
off at right angles to the gable a stretchout of the pro- 
file, as shown by P R, through the points in which 
draw the usual measuring lines. Place the J-square 
parallel to the stretchout line, or, what is the same, at 
right angles to the lines of the gable, and, bringing it 
successively against the points in the miter line D M, 
cut the corresponding measuring lines. A line traced 
through the points of intersection from 2 to 7 (that is, 
from U to V) will give the pattern for the curved por- 
tion of the profile. 

As any section of a sphere is a perfect circle whose 
length of radius depends upon the proximity of the 
cutting plane to the center of the sphere, the curves S 
to U and V to T of the pattern, representing the plain 
surfaces 1 2 and 7 S of the profile, must be arcs of 
circles, whose lengths of radius can be determined from 
the elevation. As the pattern for the plain surface 1 2 
is simply a duplicate of the cut from D to G of the 



elevation, set the dividers to the radius E G- of the ele- 
vation, and from S and U respectively as centers strike 
arcs, which will be found to intersect at N. Then N 
is the center by which to describe the arc S IT. To 
find the radius for the curve from V to T continue the 
line S M through the sphere, cutting its opposite sides 
at M and L ; then M L will be the diameter of the 




Fig. 364. — The Pattern for the Miter Between the Moldings of Adjacent 
Gables Upon a Square Shaft, Formed by Means of a Ball. 



circle of which the arc 7 S or Y T is a part. There- 
fore with K M (one-half of M L) as a radius, and Y and 
T respectively as centers, strike arcs, which will inter- 
sect in the point 0. From 0, with the same radius, 
describe the arc Y T. Then S U Y T will be the pat- 
tern of the molding to miter against the ball. 



Note.— The remaining problems in this section of the chapter involve the necessity of " raking " or de- 
veloping a new profile from the given or normal profile before the pattern for the required part can be ob- 
tained. One of the principal characteristics of this work is that, as the normal profiles are usually spaced 
into equal parts for convenience in beginning the work, the resulting or raked profiles must by force of cir- 
cumstances be made up of a number of unequal spaces ; in consequence of which their stretchouts must be 
transferred to given straight lines, space by space, as they occur upon the new profiles. 



Pattern Problems. 



177 



PROBLEM 74„ 

The Pattern of a Flaring: Article of which the Base is an Oblong: and the Top Square. 



Let A B D E of Fig. 365 be the elevation of 



the 



article, and FNOI the plan at the base, KMPL 
being the plan at the top. If the sides are to be de- 
veloped in connection with the top (supposing the 
bottom to be open) proceed for the pattern as follows : 
Draw K' M' P 1 L', Fig. 366, equal in all respects to 
K M P L of the plan. Through the center of it, and 
at right angles to each other, draw lines V U and S T 
indefinitely. While the elevation in Fig. 365 shows 
the slant hight of the ends it does not give the slant 



and L 1 P 1 , making them in length equal to F N and 
I of the plan, letting the j)omts V and U come mid- 
way of their lengths respectively. Draw K 1 F', M 1 N 1 , 
L' I' and P' O 1 , thus completing the pattern for the 
sides. Upon S T set off Z T from M 1 P 1 , and Y S 
from K' L', in length equal to A B or D E of the ele- 
vation, and through the points S and T draw F' I* and 
W 0' parallel to K" L' and M 1 P', and in length equal 
to F I and N of the plan, letting the points S and T 
fall midway of their lengths respectively. Draw F a K', 





1 

i 1 

1 

1 i 

! ! 






H G 


R 


K 

L 




^_- -- , *~ 


M 








^"~"~-— ^_ 



Fig. iitij. — Elevation and Plan. 




Fig. 3fiG.— Pattern. 
The Pattern of a Flaring Article of Which the Base is Oblong and the Top Square. 



hight or profile of the sides, therefore through the ele- 
vation, and perpendicular to the base and top, draw 
the line C G, which will measure the straight hight of 
the article. From G set off G II, in length equal to 
M R of the plan. Draw H C. Then H C will be the 
profile of the article through the side, and therefore 
the width of the pattern of that portion. Upon V U 
of the pattern, from K 1 M', set off W "V, and from L' 
P' set off X U, in length equal to II C of the eleva- 
tion. Through U and V draw lines parallel to K' M' 



F L', 1ST 2 M 1 and 2 P', which will complete the pattern 
of the ends. 

If it is required to produce the pattern with the 
sides joined to the bottom, supposing the top to be 
open, lay out first a duplicate of F N I 0, through the 
center of which draw the stretchout lines V U and S 
T as before, and proceed in the same general manner as 
described above to obtain the sides, placing their 
wider ends against corresponding sides of the base oi 
bottom. 



178 



The New Metal Worker Pattern Book. 



PROBLEM 75. 

The Envelope of the Frustum of a Pyramid which is Diamond Shape in Plan. 



In Fig. 367, let A B D E be the elevation and 
K6I0 the plan of the pyramid at the base. Project 
the points B and D into the plan, as shown, locating 
the points M and P, and draw the sides of the plan at 
top, each parallel to the corresponding line of the 
plan at the base. By projection from G or of the 
plan draw C F of the elevation, representing R of 
the plan and also the straight hight of the frustum. 

Before the slant hight or stretchout of a side can 
be obtained it will be necessary to construct a section 
on any line crossing the plan of the side at right angles 
as S T. Therefore extend the top and bottom lmes 
of the elevation, as shown dotted at the right, cutting 
the vertical line S 1 S a , thus making S 1 S 2 equal to the 
straight hight of the frustum. Upon the base line 
extended set of from S 1 the distance S 1 T 1 , equal to S T 
of the plan, and draw S* T'. Then will S 3 T' be the 
true profile or slant hight of the frustum. 

At right angles to M R of the plan draw S W, 
making its length equal to the slant hight of the frus- 
tum, as shown by S 3 T 1 of the section. Through W 
draw N H indefinitely, parallel to K 0. At right 
angles to K 0, through the points K and 0, draw 
lines K N and H, cutting N H in the points N and 
H, thus establishing its length. Connect M N and 



R H. Then MRHN will be the pattern of one of 
the four sides composing the article. 



S 1 T 1 




PATTERN. 



Fig. 867.— The Envelope of the Frustum of a Pyramid Which is 
Diamond Shape in Plan. 



PROBLEM 76. 



The Pattern of the Flaring: End of an Oblong- Tub. 



In Fig. 368, A B D C shows the elevation and 
N P O R the plan of a vessel having straight sides 
and semicircular ends, one end of which is slanting. 
First draw a correct plan and elevation of the article, 
seeing that each point of the elevation is carefully 
projected from its corresponding point in the plan. 
Divide half of the boundary line of the top into any 
number of equal spaces, commencing at 0, all as 
shown by the small figures 1, 2, 3, etc., in the plan. 
From the points thus obtained carry lines vertically 
until they cut the top of the elevation, as shown in 
the points between B and L ; also continue the lines 
downward until they meet the line T 0, all as shown. 
From the points between L and B thus obtained draw 



lines parallel to B D, producing them upward indefi- 
nitely, and continue them downward until they meet 
the bottom line of the elevation F D, as shown. At 
right angles to the lines thus drawn, and at any con- 
venient distance from the elevation, draw G H. With 
the dividers, set off from the line G H, on each of the 
lines drawn through it, the distance from T 0, on the 
lines of corresponding number, to the curved line P 0. 
In other words, make G K equal to T 6 of the plan. 
Set off spaces on the other lines corresponding to the 
distance on like lines in the plan. Through the points 
thus obtained trace a line, as shown by K H. Then 
G H K will be the half profile of the end of the 
vessel on any line, as L M, drawn at right angles to 



Pattern Problems. 



179 



the line "D B. The stretchout of the pattern is to 
be taken from the profile thus constructed. At 
right angles to D H, and at any convenient distance 
from it, draw U V, upon which lay off twice the 
stretchout of K H, numbering each way from the 
central point 1, as shown. From the points in the 
stretchout thus obtained draw measuring lines at right 



sponding measuring lines, as shown, for the top of the 
pattern. If it is desired to make a joint upon the line 
E F of the elevation, the triangular shaped piece ELF 
may be added to the pattern as follows : With the di- 
viders take the distance E F of the elevation as radius, 
and from the point Z of the pattern as center describe an 
arc. In like manner, with E L of the elevation as 











PLAN R 
Fig. SGS. — Pattern of the Flaring End of an Oblong Tub. 



angles to it indefinitely. With the blade of the 
T-square set at right angles with D B, and brought 
successively against the points in F D, cut lines of 
corresponding number drawn through the stretchout. 
Then a line traced through the points of intersection, 
as shown by Y Z, will be the pattern of the bottom 
end of the piece. In like manner bring the blade of 
the T-square against the points in L B and cut corre- 



radius, and point E in the upper line of the pattern as 
center, describe a second arc, cutting the first arc in 
the point W. Connect W with R and also with Z. 
The triangular piece at the opposite end terminating 
in point X is added in a similar manner, thus com- 
pleting the entire pattern of that portion of the vessel 
from the line E F to the right. 



PROBLEM 77. 



Pattern for the Flaring: Section of a Locomotive Boiler. 



While the pattern here described is especially 
adapted to the tapering section or " taper course " of 
a locomotive boiler its principles are equally applicable 
to tanks, cans or pipes whose shapes are governed by 
the spaces or positions which they are to occupy. The 



section of the boiler at A F, Fig. 369, is round, as 
shown by I N L M. The lower half of the circle I M 
L is the profile from L F to G D, but the upper half 
is raked or slanted from B K to C H, retaining its semi- 
circular character at C H. The line H G is a vertical 



180 



Tlte New Metal Worker Pattern Book. 



line, as shown by S L of the sectional view, and the 
surface HK6 being vertical is simply a flat triangular 
surface, exactly as shown in the elevation. 




Fig. S69. — Pattern for the Flaring Section of a Locomotive Boiler. 

Since the part B K H C is semicircular when cut 
upon any vertical line, the first step will be to obtain 
a section of it as it would appear if cut upon a line at 
right angles to B C. This section must be derived 
from the normal section of the level part, and may be 
done as follows : Assume any line, as U W, drawn at 



right angles to B C at any convenient position outside 
of the elevation, as the vertical center line of the new 
section. Divide one-half the normal section, as N L, 
into any convenient number of spaces, as shown by the 
figures, and from the points thus obtained draw lines 
parallel to A B, cutting B K, as shown, also extend- 
ing them back to the center line N M. From B K 
carry them parallel to B C, cutting the line C H, and 
extend them indefinitely, cutting also the line U W. 
With the dividers measure the horizontal distance of 
the various points in the normal profile N L from the 
center line N M, and transfer these distances to lines 
of corresponding number, measuring each time from 
IT W. Thus make W 7 equal to 7 ; the distance 
from IT W to the point 6 equal to P 6 ; and so con- 
tinue till all the distances have been measured. A 
line traced through these points will constitute a profile 

of the raking portion on aline 
at right angles to its direction, 
and B K and C H will be the 
miter lines. To develop the 
pattern first lay off the stretch- 
out of the profile TUV upon 
any line drawn at right angles 
to B 0, as A 1 B 1 . As the 
points in U T have already 
been dropped upon the miter 
lines in the previous process 
it is now only necessary to place the T-square parallel 
to A 1 B 1 and, bringing it successively against the 
points in C II and B K, drop lines cutting the 
measuring lines of corresponding number. A line 
traced through the points of intersection, as shown by 
X X Y Y, will be the pattern of the raking portion 
B K H C. To this may be added the flat triangular 
piece K H G, as shown by X Y Z. From the points 
X and Z lines may be drawn at right angles to X Z, 
as shown by Z J and X Q, extending them sufficiently 
to complete the lower portion of this part of the boiler, 
shown by K E D G of the elevation 



PROBLEM 78. 

The Pattern for a Blower for a Grate. 



In Fig. 370 D F K H E shows a front view, P L 
M O a side view, and A G B a plan of a blower. The 
conditions which determine the course to be pursued 
in arriving at the pattern are that its upper outline 



shall conform to the semicircle F K H of the- eleva- 
tion, and that it shall slant from K to G at an angle 
indicated by the line L M of the profile. Therefore, 
the first step will be to determine a true section of its 



Pattern Problems. 



181 



upper por 
L M from 



tion or hood upon a line at right angles to 
the point N of the side view; after which, 




Fig. $70. — Pattern, for a Blower for- a Grate. 

with N M and N L as miter lines, the pattern can be 
developed in the usual manner. To obtain a true pro- 



file of the hood, divide one-half of the semicircle F K 
H into any number of equal spaces, and carry lines 
from each of the several points to the vertical line L 
1ST of the side view. From the points thus obtained in 
L N carry lines parallel with L M indefinitely, which 
intersect at right angles by the line T S, located at any 
convenient point outside of the diagram. With the 
dividers take the horizontal distances between the points 
in the arc F K to the line K G, and set them off on the 
lines of corresponding number, measuring from the 
line T S. Then a line drawn through the points thus 
obtained, and as indicated by T R, will be a correct 
section through the inclined portion of the blower. 
Take the stretchout of the profile T R point by point 
and place the spaces on the line U V, which is drawn 
at right angles to L M. Through the points in U V 
draw the usual measuring lines at right angles to it. 
As the points from the profile T R have already been 
dropped upon both the miter lines M N and N L, it is 
only necessary to carry them, at right angles to L M, 
on to the measuring lines of corresponding numbers. 
Then a line traced through the points thus obtained, 
and as indicated by F 1 K 1 H 1 , will be the desired pattern. 



PROBLEM 79. 



Pattern for a Can Boss to Fit Around a Faucet. 



In Fig. 371 is shown a top and side view of a 
boss whose sides are in part parallel and just suffi- 
ciently apart to allow the faucet to fit between them. 
L 1ST represents the diameter of the opening at the tojj. 
KLIN represents the general shape of the boss 
where it joins the can and is the result of the condi- 
tions existing in the side view, but is not made iise 
of in the process of obtaining the pattern. The essen- 
tial points are the curve of the can body, D A B E, 
the diameter of boss at top, LON, the distance be- 
tween D and E and the distance X C, all of which are 
shown in the side view. 

Divide one-quarter of the plan of the top, as indi- 
cated by N, into any convenient number of spaces, 



as indicated by 1, 2, 3, etc. From the points thus 
established drop lines vertically, cutting the line repre- 
senting the top in the side view, as shown from F to' 
C. From the points thus established in F G carry 
lines parallel to the side F D, producing them until 
they cut the curved line D A B E, as shown between 
D and A. The next step to be taken is to obtain the 
profile which would be shown by a section taken 
through the article at right angles to the line D F. 
For this purpose at any convenient point draw a line 
through D F and at right angles to it, as shown by P R. 
From the points established in the plan of the top, as 
shown from to N, carry lines vertically until they 
meet the horizontal line K M passing through the een- 



182 



Tlie New Metal Worker Pattern Booh. 



ter of the top, as shewn. Taking the length of each 
of the distances thus obtained in the dividers, set it 
off from either side of P B on the lines of correspond- 
ing numbers, and through the points thus obtained 
trace the curve, as shown. Then this curve will repre- 
sent the required section from which the stretchout of 
the envelope may be obtained On the line E P, pro- 
duced sufficiently outside of the side view for the pur- 
pose, lay off the stretchout of one-half of this curve, 
as shown, and through the points thus established 
draw measuring lines parallel to D F. Then, with 
the T-square placed parallel to P E, or, what is the 
same, at right angles to D F, and brought successively 
against the points in the profile of the can body be- 
tween D and A, cut the measuring lines of correspond- 
ing numbers. In like manner bring the T-square 
against the points in the top of the article shown from 
F to C and cut the measuring lines of corresponding 
numbers. Then lines traced through the points thus 
obtained, as shown by D' A' and F' C, will be one- 
half of the pattern of one of the ends. As that por- 
tion of the boss lying between points A, B and C is 
simply a flat triangular piece it is only necessary to 
add a duplicate of its shape to that part of the pattern 
just obtained, bringing one of its straight sides against 
the line 5, all as shown. To the other straight side 
0' B' must be added a duplicate of the first part of the 
pattern reversed, as shown by B' C G' E'; the result- 
ing shape will than constitute the pattern of one-half 
the boss. 




Fig. S71. — Pattern for a Can Boss to Fit Around a Faucet. 



PROBLEM 80. 



The Patterns for a Molded Base in which the Projection of the Sides is Different from that of the Ends. 



Let A B C D, in Fig. 372, represent the side view 
and E F G H the plan of a base in which the projec- 
tion of the sides, as shown at P, is less than that of 
the ends, as shown at M C. B C and A D show the 
profile of the ends of the base. As the projection 
through the sides of the base is less than that of the 
ends, a profile must be obtained through the side P 
in plan, from which to obtain the stretchout in pi-o- 
ducing the pattern of sides. To obtain the pattern for 
the end proceed as follows : Divide the profile B C 
into an equal number of parts, as shown by the small 
figures, and from the points obtained drop lines at right 



angles to A B until they intersect the miter lines J P 
and L G in plan. At right angles to F G draw the line 
B 2 C~, upon which place the stretchout of the profile 
B C, as shown by the small figures, through which draw 
the usual measuring lines, which intersect with lines 
of corresponding numbers drawn from the miter lines 
at right angles to F G. A line traced through these 
intersections, as shown by J' 1/ G' F', will be the re- 
quired pattern for the end of the base. To obtain the 
profile through the side proceed as follows : From B 
in elevation draw the vertical line B M, as shown, and 
from the divisions on the profile B C draw lines par- 



Pattern Problems. 



183 



allel to the lines of the moldings until they intersect 
the line B M, as shown by the small figures. From 
the intersections obtained on the miter line L G in plan, 
as before explained, draw lines parallel to H G and ex- 
tend them indefinitely, cutting the miter line K H, as 
shown. Upon K L of the plan extended set off B' M', 



the side of the base. To obtain the pattern for the 
side proceed as follows : At right angles to H G draw 
the line B 3 ISP, upon which place the stretchout of the 
profile B' N', as shown by the small figures. It will be: 
noticed that the spaces in the profile B' 1ST' are unequal, 
and therefore each must be separately placed on the line 







Fig. S72. — The Patterns for a Molded Base in Which the Projection of the Sides is Different from that of the Ends. 



in hight equal to B M of elevation, and transfer the 
spaces from B M to B' M', as shown. At right angles 
to B' M' and from the points on same draw lines, as 
shown, intersecting lines of corresponding numbers 
drawn from the miter line L G. A line traced through 
these intersections will be the desired profile through 



B 3 1SP: Through the points in thisstretchout line draw 
the usual measuring lines, as shown^ which intersect 
with lines of corresponding numbers drawn at right 
angles to II G from the miter lines K H and L G. A 
line traced through these intersections, as shown by 
K' H' G' L', will be the desired pattern for the side. 



PROBLEM 81. 



The Patterns for an Elliptical Vase Constructed in Twelve Pieces. 



The first essential in beginning the work is an 
ellipse, which may be drawn by whatever rule is most, 
convenient, and which must be of the length and 
breadth which the vase is required to have. Draw the 
plan of the sides of the vase about the curve, as shown 



in Fig. 373, in such a manner that-all the points X, 1% 
Z, etc., shall have the same projection beyond the 
curve. Complete at least one-fourth of the plan by 
drawing miter lines, as shown by P C, M C, C, U G 
and N C. Above the plan construct an elevation ->£ 



184 



The New Metal Worker Pattern Book. 



the article, as shown by H L K G. Only the profile 
H V W L of the elevation is needed for the purpose of 



considered before the article is constructed. As the 
projection of each of the sides upon the plan is dift'er- 




Fig. 873.— The Patterns for an Elliptical Vase Constructed in Twelve Pieces. 



pattern cutting, but the other lines are desirable in 
process of designing, in order that the effect may be 



ent when measured from the center C on lines at right 
angles to the lines of the sides, it will be necessary 



Pattern Problems. 



185 



first to develop the profiles of each of the varying sides 
from the normal profile II V W L. Therefore, divide 
II V W L in the usual manner, and from the several 
points in it drop lines across its corresponding section 
(No. 1) of the plan. 

Across the second section in the plan, from the 
points already obtained in U C, draw lines parallel to 
O II, the side of it cutting C, and produce them 
until they meet A C, which is drawn from C at right 
angles to U produced. Then the points in A C give 
the jjrojections from which to obtain a profile of the 
section numbered 2. In like manner continue the 
points from C across the third section in the plan, 
parallel to M, the side of it cutting M C, and pro- 
duce them until they cut C B, which is drawn from C 
at right angles to M produced. Then C B contains 
the points requisite in obtaining a profile of the third 
section. Continue the points in C M across the fourth 
section, cutting its other miter line C P. From C 
draw C D at right angles to the side P M of the sec- 
tion, cutting the lines drawn across section 4. Then 
upon C D will be found the points necessary to 
determine the profile of the fourth pattern. Pro- 
duce the line of the base of the elevation indefinitely, 
as shown by C C" C 3 , and also the line of the top A' 
B 2 D 3 . From the several points in the profile H V W 
L draw lines indefinitely, parallel to the lines just de- 
scribed and as shown in the diagram. From C, upon 
the base line produced, set off points corresponding to 



the points in C A of the plan, making the distance 
from C in each instance the same as the distance from 
C in the plan. Number the points to correspond with 
the numbers given to the points in the profile HTf 
L, from which they were derived. In like manner 
from C 2 set off points corresponding to the points in C 
B of the plan, numbering them as above described. 
From C 3 set off points corresponding to those in CD of 
the plan, likewise identifying them by figures in order 
to facilitate the next operation. From C erect the 
perpendicular C A 1 ; likewise from C 2 and C 3 erect the 
perpendiculars C 2 B 2 and C 3 D 3 . From each of the 
points laid off from C, and also from each of those laid 
off from C 2 and C 3 , erect a perpendicular, producing it 
until it meets the horizontal line drawn from the profile 
H V W L of corresponding number. Then lines traced 
; through these several intersections will complete the 
profiles, as shown. Perpendicular to the side of each 
section in the plan lay off a stretchout taken from the 
profile corresponding to it, just described, and through 
the points in the stretchouts draw measuring lines in 
the usual manner, all as shown by E F, E 1 F' E 2 F 2 and 
E 3 F 3 . Place the "]"-scpiare parallel to each of these 
stretchout lines in turn, and, bringing it against the 
several points in the miter lines bounding the sections 
of the plan to which they correspond, cut the measur- 
ing lines in the usual manner. Then lines traced 
through the points of intersection thus obtained, all as 
shown in the diagram, will complete the patterns. 



PROBLEM 82. 



The Patterns for a Finial, the Plan of which is an Irr-egular Polygon. 



In the central portion of Fig. 374 is shown the 
plan B C D E F G H, upon which it is required to 
construct a finial, the only other view given being a 
section through one or the sides, being that numbered 
1 on the plan. 

The section of side ABC, or No. 1, is shown 
above and in line with the plan of the same, and is 
marked Profile No. 1, and is a section on the line A M, 
which is drawn at right angles to B C of the plan. 
To obtain the pattern of A B C of plan, or No. 1, 
divide the profile K L in the usual manner, and, with 



the T-square placed parallel with C B of plan and 
brought successively against the several points in 
profile K L, drop lines cutting the miter lines A B 
and A C. 

On A M extended, as Al Ml, lay off a stretchout 
of K L of profile, through the points in which draw 
the customary measuring lines. Place the T-square 
parallel to the stretchout line Al Ml, and, brinaino- it 
against the several points in the miter lines A B and 
A C, cut corresponding measuring lines. Tracing 
lines through the points thus obtained, as shown bv 



186 



The New Metal Worker Pattern Book. 



Al CI Bl, will give tlie pattern of part of article 
shown on plan by A B C. 



at right angles to each — that is, since AN, AO and 
A P, drawn at right angles respectively to D, D E 



PROFILE 
NO. 4 



PROFILE 
NO. 3 



PROFILE 
NO.1 



PROFILE 
NO. 2 




Fig. 374.— Pattern for a Finial, the Plan of Which is an Irregular Polygon. 



Since the point A in the plan is not equidistant 
from all the sides of the same, when measured on lines 



and E F, differ in length from each other and from 
A M a correct section must be obtained for each 



Pattern Problems. 



187 



of the other sides before their patterns can be 
developed. 

These different sections can be most conveniently 
obtained at one operation in the following manner : 

"With the T-square placed parallel with D C, and 
brought successively against the points in A C, draw 
lines cutting A D and A N. Then the points in AN 
can be used to obtain a profile of section No. 2. Also 
continue the points from A D across the third section 
of the plan, and parallel with D E, and produce them 
until they cut A E, also A 0, which is drawn from A 
at right angles to D O produced. Then A contains 
the points necessary in obtaining a profile of the third 
section. Continue the points from A E across the 
fourth section of the plan, and parallel with E F, cut- 
ting A P and A F. Then the points in A P can be 
used to obtain a profile of section No. 4. 

"While the projection of the several points in each 
of the new profiles can be obtained respectively from 
the lines A N, A and A P, the hights of the several 
points must be the same in all and must be derived 
from the normal profile K L. Therefore continue 
J L, which represents a horizontal line of profile No. 1 , 
in either direction, as shown by L4 L2. From the 
several points in profile No. 1 draw lines parallel with 
L4 L2, extending them indefinitely in either direction. 
At any convenient position on L4 L2 set off points 
corresponding to the points in A N of the plan, as 
shown at J2 L2, numbering them to correspond with 



the points in A N and in the normal profile. From 
each of the points in J2 L2 erect lines perpendicular 
to the same, intersecting lines of corresponding number 
drawn from K L. Then a line traced through the points 
of intersection, as shown from K2 to L2, will be the cor- 
rect section on A N of the plan, from which to obtain 
a stretchout of piece No. 2. 

The sections of pieces No. 3 and No. 4 are ob- 
tained in a similar manner from A and A P. J3 
L3 is a duplicate of A and J4 34 of A P. Perpen- 
diculars are erected from each of the points cutting 
horizontal lines of corresponding number, thus devel- 
oping K3 L3 and K4 L4. 

To obtain the pattern of A D C (No. 2) continue 
A N downward indefinitely, upon which lay off a 
stretchout of K2 L2, as shown by A2 N2, through 
the points in which draw the usual measuring lines. 
Place the J-square parallel with A2 N2, and, bringing 
it against the several points in A C and A D, cut 
measuring lines of corresponding numbers. Lines 
traced through these points, as shown by A2 C2 and 
by A2 D2, will be the pattern sought. 

The stretchouts for pieces Nos. 3 and 4 are taken 
respectively from profiles 3 and 4. A3 03 is the 
stretchout of K3 L3 and is laid off on a continuation 
of A 0, while A4 P4 is taken from K4 L4 and is set 
off on a continuation of A P. The remaining opera- 
tions are the same as those employed in obtaining the 
other pieces. 



PROBLEM 83. 



Pattern for a Three-Piece Elbow, the Middle Piece Being: a Gore. 



Let A B C D E F G in Fig. 375 be the elevation 
of a three-piece elbow to any given angle, as G F E, 
the middle piece of which, BCH, forms a gore extend- 
ing around one-half the diameter. The lines H B and 
H C are drawn parallel respectively to the ends of the 
two outer pieces, therefore the patterns for the end 
pieces will be straight from H to C and H to B and 
mitered from H to F. To obtain the pattern for one 
of the ends, as F H CDE, divide N K L into any 
convenient number of equal parts. "With the J-square 
at right angles to N L, carry lines from the points in 



N K L, cutting the miter line F H C, as shown. On 
any line, as E D extended, lay off a stretchout of N 
K L, as shown by e d, through the points in which 
draw the usual measuring lines, as shown. With the 
T-square brought successively against the points in F 
H C, cut corresponding measuring lines, as shown. A 
line traced through the points of intersection, as shown 
bj/h" c, will be the half pattern of end, as represented 
in elevation by F H C D E, or in profile by N K L. 
The other half of the pattern can be obtained by dupli- 
cation. 



188 



Tire New Metal Worker Pattern Book. 



Since H C is drawn parallel to E D, the distance 
H J is less than one-half the diameter of the normal 
profile, or less than L, therefore it will be necessary 
before obtaining the pattern of B H C to obtain a cor- 
rect section on line H J of elevation. To do this 
first place the T-square parallel with B C and carry lines 
from the points in H C, cutting H J and II B. ISText 
draw any line, as K' M' of the section, and erect the 
perpendicular H' J'. From II', on H' J', set off the 
spaces in H J of elevation, transferring them point by 
point. Through the points thus obtained draw lines 
parallel with K' M', as shown. With the dividers take 
the distance across K L or L M, on the several 
lines drawn parallel with K M, and set off the same 
distance on lines of corresponding number drawn 
throuo-h H' J'. Thus H' M' and H' K' are the same as 
K and M. A. line traced through the points thus 
obtained, as shown by K' J' M', will be the section 
desired. 

For the pattern of HBC, lay off on HJ extended, 
as h h', a stretchout of K' J' M' of section, through which 
draw the usual measuring lines. With the T-square 
placed parallel with H J, and brought successively 
against the points in B H and C H, cut corresponding 
measuring lines drawn through h h', as indicated by the 
dotted lines. Lines drawn through these points of 
intersection, as shown by h b h' c, will be the pattern 
of the part shown in elevation by B II G. 




H' 

SECTION 



Fig. S75.— Pattern for a Three-Piece Elbow, the Middle Piece 
Being a Gore. 



PROBLEM 84. 



The Patterns of a Tapering Article which is Square at the Base and Octagonal at the Top. 



A B D C in Fig. 376 shows the plan of the article 
at the base, IKLMHGFE represents the shape at 
the top, E 3 H 2 D 2 C 3 is an elevation of one side. In 
order to obtain the slant night of the octagonal sides it 
will be necessary to construct a diagonal section or ele- 
vation. Therefore extend the lines of the base C 3 D 2 and 
top E 3 H 2 , as shown, to the left, through which draw a 
vertical line, as B 1 C\ Upon the line of the base set 
off each way from C 2 a distance equal to A E or E D 
of the plan. In like manner upon the line of the top, 
set off from B 1 each way a distance equal to one-half 



I G. Draw V A' and G' D 1 , thus completing a diagonal 
section. If it is desired to complete a diagonal eleva- 
tion, set off E 2 F 2 equal to E F of the plan and draw 
lines to C 2 , as shown. 

To obtain the pattern of one of the smaller sides, 
produce the diagonal line B C, upon which set off the 
length or stretchout of I 1 A"', as shown by N C, and 
draw the measuring line E 1 F 1 . By means of the 
T-square, as indicated by the dotted lines, set off E 1 F' 
equal to E F of the plan and draw C E 1 and C F 1 . 
Then E' C F 1 is the pattern of one of the smaller sides 



Pattern Problems. 



189 



Fig. 377.— Pattern in One Piece. 



For the pattern of one of the 



larger 



of the article 

sides, draw R P perpendicular to the side A C, upon 
which set off P, in length equal to E 3 C 3 of the ele- 
vation, at right angles to which through and P draw 
measuring lines. 

By means of the J-square, as shown by the dotted 
lines, make A 2 C 4 equal to A C of the plan. In like 
manner make F E 4 equal to I E of the plan. Connect 
A 5 F and C 4 E 4 . Then A 2 F E 4 C 4 will be the pattern 
of one of the larger sides of the article. If for anv 
reason the pattern is desired to be all in one piece the 
shapes of the different sides may be laid off adjacent 
to each other, the large and small sides alternating, all 
as indicated by i i' a a 1 , Fig. 377. 




Fig. 376.— Elevations, Plan and Patterns. 
Tapering Article Which is Square at the Base and Octagonal at the Top. 



PROBLEM 85 

The Patterns of a Finial, the Plan of which is Octagon with Alternate Long and Short Sides. 



In Fig. 37S, let A L M N P R S T be the ele- 
vation of the finial corresponding to the plan which is 
shown immediately below it. The elevation is so 
drawn as to show the profile of one of the long sides, 
for the pattern of which proceed in the usual manner. 
Divide the profile A L M N P into any number of 
convenient spaces, as shown by the small figures, and 
from the points thus obtained drop lines across the 
corresponding section in the plan, cutting the miter 
lines DEC and D" F C, as shown. A duplicate of the 
part E C F of the plan is shown below by E 1 C 2 F', and 
in the demonstration C 2 E 1 and C 2 F 1 are to be consid- 
ered the same in all respects as C E and C F. The 
same may be said of the two parts of the stretchout 
line bearing like letters; the division having been 



made on account of the extreme length which the pat- 
tern would have if made in one piece. Perpendicular 
to D D 2 lay off a stretchout, as shown by G H, through 
the points in which draw measuring lines in the usual 
manner. Place the J-square parallel to the stretchout 
line, and, bringing it against each of the several points 
in D E C and D 2 F C, cut the corresponding measuring 
lines. Then a line traced through these points of 
intersection will be the pattern sought. 

For the pattern of the short sides a somewhat dif- 
ferent course is to be pursued. As the distance from 
C to the line K E is greater than that from C to E F a 
profile of the piece as it would appear if cut on the line 
C D must first be obtained. To do this proceed as 
follows : From the points in C E, dropped from the 



190 



Tlie New Metal Worker Pattern Book, 




Fig. 378.— Elevation, Plan and Patterns. 
The Patterns of a Finial, the Plan of Which is Octagon with Alternate Long and Short Sides. 



31+7 4 '6 
B' 5 , 7 

Fig. 379.— Diagonal Section. 



profile, carry lines parallel to E K across D, cutting 
C K, as shown. At any convenient place lay off B 1 
P', Fig. 379, in length equal to C D of the plan, on 



which lay off points corresponding to the points ob- 
tained in C D, and for convenience in the succeeding 
operations number them to correspond with the nam- 



Pattern Problems. 



191 



bers in the profile from which they are derived. At 
B 1 erect the perpendicular B' A', equal to B A of the 
elevation. From the several points in the profile of 
the elevation draw horizontal lines cutting the central 
vertical line A B, as shown. Setoff points in A 1 B 1 in 
Fig. 379 to correspond, and through these points draw 
horizontal lines, which number for convenience of iden- 
tification in the following steps. From the several points 
in B 1 P 1 carry lines vertically, intersecting correspond- 
ing horizontal lines. Then a line traced through these 
points, as shown by A 1 L' M' N ! O 1 P\ will be the 
profile of the short side on the line C D of the plan. 



After obtaining the profile as here described, for the 
pattern of the short side proceed as follows : Perpen- 
dicular to K E of the short side, or on C D extended, 
lay off a stretchout of the diagonal section A' O 1 of 
Fig. 379, as shown by C 1 D', through the points in 
which draw measuring lines in the usual manner. Place 
the T-square parallel to the stretchout line, and, bring- 
ing it against the several points in the miter lines D K 
C and DEC bounding the short side in the plan, cut 
the corresponding measuring lines. Then a line traced 
through these points, as shown in the diagram, will be 
the required pattern. 



PROBLEM 86. 

The Pattern for a Gore Piece Forming a Transition from an Octagon to a Square, as at the 

End of a Chamfer. 



In Fig. 380, let F F F F represent the plan of the 
square portion of a shaft and A A A A that of the 



Pattern 




Fig. 380.— The Pattern for a Gore Piece Forming the Transition 
from an Octagon to a Square. 

octagon portion. Let D P C be the elevation of the 
gore piece which is required to form the transition be- 
tween the two shaDes. The outline C D, which repre- 



sents the intersection of the gore piece with the side 
of the shaft, may be of any contour whatever at the 
pleasure of the designer, the method of laying out the 
pattern being the same no matter what its outline. By 
reference to the plan it will be seen that the lines of 
the molding, of which C D shows only the termination, 
run octogonally, or in the direction of A A. There- 
fore, before a stretchout of the piece can be obtained 
a correct profile must be developed on a line at right 
angles to its lines — that is, on the line E F. To do this 
proceed as follows : Divide the line C D, as it appears 
in the elevation, into any convenient number of spaces, 
as shown by 1, 2, 3, i, etc. From the points thus 
obtained drop lines down ttpon the side of the plan 
A F, which should be placed in line below the ele- 
vation. 

Continue the lines from the side A F across the 
corner, as shown, all parallel to A A of the octagon, 
crossing E F, and number the lines to correspond 
with the numbers of the points in the elevation from 
which they were derived. Draw the vertical line 
G H at a convenient distance from P D, and cut G H 
by lines drawn at right angles to it from the points in 
C D, as shown by the connecting dotted lines. G II 
then may be considered to represent the point F in the 
plan, or 11 of the numbers on the line E F. From 
G H, on each of the several lines drawn through it, lay 
off a distance equal to the space from E to the corre- 
sponding number in the same plan. Thus lay off from 
Gr H on line 1 a distance equal to 11 1 on E F, and 
on line 2 a distance equal to 11 2 of E F, and so on 
for each of the lines through G H. Then a line traced 
through these points, as shown by I H, will be the 



192 



The New Metal Worker Pattern Book. 



profile of the gore piece, or trie shape of its section 
when cut by the line E F. 

Prolong E F, as shown by K L, and lay off on 
the latter a stretchout of the profile I II, the spaces of 
which must be taken from point to point as they occur, 
so as to have points in the stretchout corresponding to 
the points on the miter lines A F, previously derived 



from C D. Tnrough the points thus obtained draw 
the usual measuring lines, as shown. Place the 
T-square at right angles to the measuring lines, or, what 
is the same, parallel to E F, and, bringing it against 
the points in A F and F A, cut the corresponding lines 
drawn through the stretchout. Lines traced through 
these points, as shown, will constitute the pattern. 



PROBLEM 87. 

The Pattern for a Gore Piece in a Molded Article, Forming a Transition from a Square to an Octagon. 

In Fig. 381, let A B D C represent the elevation 
ol an article of which Cr H I J is the half plan at the 
base and K L M NOP the half plan at the top. A C 
of the elevation is the normal profile or profile of one 
of the square sides, and L II and M II of the plan 
show the miter lines between the square sides and the 
gore piece. C E and D F, the elevations of the miter 
lines II M and I N, are shown as part of the design, 
but are not necessary in cutting the pattern. 

As only the normal profile, which would be used 
in cutting the pattern of one of the square sides, is 
shown in the elevation, the first step will be to obtain 
from this a profile of the gore piece, or in other words, 
a section upon its center line, R H of the plan. Divide 
the profile A C into any convenient number of parts, 
and from the points obtained drop lines at right angles 
to A B, cutting the miter line L II in plan, as shown. 
From the intersections obtained on the miter line L H 
draw lines parallel to L M, as shown, cutting the other 
miter line II M, and continue them indefinitely. At 
any convenient position outside the plan draw the line 
A 1 A 2 parallel to H R, and draw a duplicate of the 
profile A C in the same relative position to A 1 A 2 that 
A G holds to A B, and divide the same into the same 
spaces as A C, all as shown by A 1 0. From the points in 
A 1 C draw lines parallel to A 1 A 2 , cutting lines of 
corresponding number drawn through the plan of the 
gore piece. A line traced through these intersections, as 
shown from A 2 to C 2 , will be the profile of the transition 
piece from which to obtain the stretchout for the pattern. 

To obtain the pattern proceed as follows : Upon 
R 1 R 2 , a continuation of H R, place the stretchout of 
the profile A 2 C 2 , as shown by the small figures, through 
which draw the measuring lines, as shown. These are 
now to be intersected by lines drawn from points of 
corresponding number upon the miter lines H L and 
H M. Lines traced through the points of intersection, 

01 Fig SSI. — The Pattern for a Gore Piece in a Molded Article, harming 

shown by R T and R S, will give the desired pattern. ] a Transition from a Square to an Octagon. 




PROFILE OF 
GORE PIECE. 



Pattern Problems. 

PROBLEM 88. 



193 



The Patterns for a Raking 1 Bracket. 



This is one of the many instances which calls for 
special draftsmanship on the part of the pattern cutter. 
Frequently the architect's drawings give only a detail 
of a bracket for the level cornice of a building:, while 
the scale elevations show one or more of the gables to 
be finished with raking brackets. In such cases the 
detail of the " level " bracket and the pitch of the roof 
are the only available facts from which to produce the 
required bracket. 

In Fig. 382, let M X or P be drawn at the re- 
quired angle, with reference to any horizontal line, to 
represent the pitch of the gable cornice. The first step 
is to redraw the normal side elevation of the level 
bracket so that its vertical lines shall be at right angles 
to the lines of the rake, all as shown at L Q P. Next, 
at any convenient distance from this draw two vertical 
lines, as M and N 1 P 1 , the horizontal distance between 
which shall be the required face width of the bracket. 
Lines projected parallel to the rake from the various 
angles in the profile between these vertical lines will 
complete the front elevation of the raking bracket. The 
additional lines E G and F H representing the sink in 
the face, A C showing the depth of the panel in side, 
and TJ D giving the depth of the sink in the face, will 
be understood from the drawing. 

To construct a side view of the raking bracket, or, 
what is the same thing, the pattern for the side (includ- 
ing the bottom of the sunken panel and the sink strips 
U D Z in the face), all hights must be measured upon 
one of the vertical lines of the face view, as M 0. To 
avoid confusion, however, and make room for other 
patterns, another vertical line, X 1 P 2 , will serve as well. 
Divide the curved portions IT to P of the face of the 
normal profile into any convenient number of small 
spaces for use in this and subsequent parts of the oper- 
ation. From all the points in the profile of face carry 
lines parallel to the rake through the side view and 
continue them till they intersect the vertical line X' P\ 
From the points thus obtained in the line X 1 P 2 carry 
lines indefinitely horizontally, as indicated. Upon 
each of the lines so drawn lay off from the line X 1 P 2 a 
distance or distances equal to the distance or distances 
upon the corresponding lines drawn across the normal 
side of the bracket. Through the points thus obtained 
trace lines, which will give the several shapes in the 
sides of the brackets corresponding to the shapes shown 



in the side of the normal bracket. It may be necessary 
to introduce in the several profiles of the normal bracket 
other points than those derived from spacing the profile. 
Use as many such points as may be necessary to deter- 
mine the position of all points in the side being con- 
structed. Then X 1 N 2 P 2 will be the pattern of the side 
of the bracket, and U 2 Z 2 D 2 will be the pattern of the 
strip forming the sides of the sink shown in the face by 
EFHG, and b l a 1 d l c 1 will be the shape of the panel in 
the side of the bracket. 

For the patterns of the several pieces forming the 
face of the bracket the profiles are to be found in the 
normal side view, from which stretchouts can be ob- 
tained when wanted, and laid out at right angles to the 
lines of the rake ; while the miter lines of any part are 
the vertical lines of the face view corresponding to that 
part of the profile under consideration. 

For the strip PEGS, forming that part of the 
face at the side of the sink, lay off a stretchout of its 
profile U Z at right angles to the lines of the rake, as 
shown by u* z\ through the points in which the usual 
measuring lines are drawn. Drop the points from the 
profile to the miter lines E S and E G ; then, with the 
T-square placed at right angles to the lines of the rake, 
and brought successively against the points in E S and 
E G, the corresponding measuring lines are cut. Then 
lines traced through these points of intersection, as 
shown by E 1 S 1 and E' G 3 , form the pattern for that 
piece. 

For the piece forming the face of the bracket below 
the sink, as shown in the elevation by S P 1 Z 1 , pro- 
ceed in like manner. A stretchout of its profile, as 
indicated by D P, is laid off at right angles to the lines 
of the rake, through which the usual measuring lines 
are drawn. The points in D P are then carried parallel 
to the rake, cutting the miter lines S and Z' P'. 
The T-square is then placed at right angles to the lines 
of the rake, and brought against the several points in 
the sides S and Z 1 P 1 , by which the corresponding 
measurina; lines are cut. In like manner it is brought 
against the points G and H, by which the shape of the 
part extending up to meet the sink is determined. 
Then lines traced through these several points of inter- 
section, as shown by H 3 Z 3 P 3 0' S 2 G 4 , form the pattern 
for that part of the face of the bracket. The upper 
part of the face of the Dracket, shown in the face view 



194 



Tlie New Metal Worker Pattern Booh 



by N 1 IT' E M, being a flat surface, as indicated in the 
side view N U, is obtained by pricking directly from 
the face view of the bracket, no development of it 
being necessary. 

To avoid confusion of lines, the sink piece E F H 



profile, as shown by u' d% is laid off at right angles to 
the lines of the rake, and through the points in it the 
usual measuring lines are drawn. The T-square is then 
placed at right angles to the lines of the rake, and, be- 
ing brought successively against the points in the sides 




Fig. 084. — Lower Return of 
Bracket Head. 



Fig. SS2. — The Patterns for a Raking Bracket. 



G is transferred to the right, as shown, by E 1 F 1 H 1 G ! . 
The profile of it, as indicated in the side view by U D, 
is divided into any convenient number of spaces, 
and through the points lines are drawn, cutting the 
miter lines E 1 G 1 and F 1 H'. The stretchout of this 



E 1 G 1 and F 1 H', the -corresponding measuring lines are 
cut. Then lines traced through these points of inter- 
section, as shown by E 2 G" F 3 H", constitute the pattern 
of the bottom of the sink. 

Of the strips bounding the panel of the side in the 



Pattern Problems. 



195 



bracket, the piece corresponding to b c in tlie side 
view, being vertical, is obtained by pricking directly 
from its elevation in tlie face view of the bracket, A B 
D 1 C being the shape. For the other straight strip 
bounding this panel, shown in the side view by a b, 
the length is laid off equal to a b, while the width is 
taken from the face view, equal to the space indicated 
by A B. For the strip representing the irregular part 
a to c proceed as follows : Divide the profile a d c into 
any convenient number of parts, from the joints in 
which carry lines crossing the face view of the same 
part, as indicated by A B D' C. At right angles to 
the lines of the rake lay off a stretchout of the profile 
just named, as indicated by a" c 3 , through the points 
in which draw the usual measuring lines. Place the 
T-square at right angles to the lines of rake, and, 
bringing it against the several points in the line A C 
and B D', cut the corresponding measuring lines drawn 
through the stretchout. Then lines traced through 
the several points of intersection thus formed, as indi- 
cated by A 1 C 3 and B' D 3 , will be the pattern of the 
curved strip forming part of the boundary of the panel 
in the side view of the bracket. 

Of the three pieces of molding forming the head 
of the bracket, the profile of the piece across the face 
is normal, as shown at L N, while that of the two side 
pieces, or returns, requires to be modified or raked 
before a square miter with the face piece can be 
effected. These principles will be further explained 
in Problems 91 to 94. The first step will be to draw a 
correct elevation of the head, which includes raking the 
profiles of the upper and lower returns. 

Divide the normal profile L N into convenient 
spaces, and from the points thus obtained carry lines 
indefinitely parallel to the rake across the top of the 
face view of the bracket. Draw duplicates of the nor- 
mal profile, placing them in a vertical position directly 
above where the new sides are required to be, as shown 
by n I and k m. Divide these two profiles into the 
same number of parts employed in dividing the normal 
profile, and from these points drop lines vertically, 
intersecting those drawn from L N". Then lines traced 
through these points of intersection, as shown by 
L 1 1ST 1 and K M, will be respectively the profiles of the 
moldings on the upper side and on the lower side of the 
bracket. Lay off a stretchout of the profile LN at 
right angles to the line of the rake and through the 
points in it draw the usual measuring lines. With the 
blade of the T-square at right angles to the lines of 
the rake, and brought successively against the several 



points in the profile N' L 1 and K M, cut the measuring 
lines drawn through the stretchout. Then lines traced 
through the points of intersection thus obtained, as 
shown by L. 3 ISP and K 1 M', will be the shape of the 
ends of the molding forming the front of the bracket 
head. 

Before laying out the pattern for the return mold- 
ing forming the upper side of the bracket head a cor- 
rect side elevation of it must be drawn. A duplicate 
of the profile L 1 1ST 1 is transferred to any convenient 
place, as shown at 1/ N 4 in Fig. 383, and parallel lines 
from its angles are extended to ihe right, as shown, 
making L 3 Q 1 equal to L Q of the side view of the 
bracket. 

At Q 1 repeat the outline L 3 N 4 , which represents 
the intersection of the bracket head with the bed mold 
of the cornice. L' N 1 of Fig. 382 is then the cor- 
rect profile and the lines L 3 1ST 4 and Q' X 3 are the miter 
lines of this return ; however, as both the miter lines 
are identical with the profile, the stretchout q x may 
be taken from either one, the other being divided into 
the same number of spaces as the first, which is easier 
than dropping the points from one to the other. The 
T-square may then be placed at right angles to the 
lines in the molding and brought successively against 
the points in the lines L 3 1ST 4 and Q 1 X 2 , and the cor- 
responding measuring lines intersected. Then lines 
traced through these points, as shown by L 4 N 6 and 
Q 3 X 3 , will form the pattern. 

The pattern for the return molding of the head 
occurring on the lower side of the bracket is obtained 
in the same manner. A duplicate of the profile K M 
of the face view of the bracket is drawn at any con- 
venient place, as shown by K 3 M 3 in Fig. 38-4. The 
proper length is given to the molding by measuring 
upon the side view of the bracket, and a duplicate of 
the profile is drawn at the opposite end. Space the 
profile K 3 M 3 into any convenient number of parts, as 
indicated by the small figures, and in like manner di- 
vide the profile K s M 3 into the same number of parts. 
At right angles to the line of the molding lay off a 
stretchout of these profiles, as shown by k 1 in', through 
which draw the usual measuring lines. With the blade 
of the T-square at right angles to the lines of the 
molding, and brought successively against the several 
points in the profiles K 3 M 3 and K 3 M 3 , cut the corre- 
sponding measuring lines. Then a line traced through 
these points of intersection, as shown by K 6 M 5 and 
K 4 M 4 , will constitute the pattern of the return mold- 
ing, or the lower side of the bracket. 



196 



Tlie Neiu Metal Worker Pattern Book. 



PROBLEM 89. 



The Pattern for a Raised Panel on the Face of a Raking; Bracket. 



In the solution of the problem stated above, and 
which is given in Fig. 385, the first requisite is the 
design or outline of the side of the normal bracket, as 
such an outline is really a section through the raking 
bracket upon a line at right angles to the rake. NST 
shows the side view of a normal bracket, or the bracket 
as it would appear in a level cornice, the part from G 
to H being molded as shown by the shaded profile, 
which profile, being a section on line a b of the normal 
bracket, is given complete at J and called the nor- 
mal profile. The first step is to derive from these 
factors a front elevation of the molded panel upon the 
face of the raking bracket. To accomplish this first 
divide the profile of the panel molding into any con- 
venient number of equal parts, as shown in the section 
shaded in the side of the normal bracket, and through 
these points draw lines parallel to the face of the 
bracket, producing them until they cut the upper sur- 
face against which the panel terminates, and in the 
opposite direction until they meet the vertical surface 
in the lower part of the bracket against which the panel 
terminates at the bottom. From the points thus ob- 
tained in the horizontal surface near the top of the 
bracket and in the vertical surface near the bottom of 
the bracket draw lines at right angles to the face, thus 
transferring the points to the line representing the outer 
face of the panel, as shown from G to II. 

These points will be used a little later in develop- 
ing the view of the panel at right angles to the face. 
Next, from the points already obtained in the line rep- 
resenting the vertical surface near the bottom of the 
bracket carry lines parallel with the rake, extending 
them across the front elevation of the bracket. In the 
diagram, to avoid confusion, these lines terminate at 
the intersections shown from A to B, but in actual 
work they would be extended across the front eleva- 
tion, thereby making also the intersections shown from 
C to D. At any convenient place in line with the 
front elevation of the raking bracket draw the normal 
profile, as shown below the elevation, and divide it 
into spaces corresponding to the spaces used in dividing 
the profile in the side view. From the points thus 
obtained carry lines vertically, intersecting those just 
drawn from the side of the normal bracket across the 
front elevation. A line traced through the points of 



intersection gives the outlines shown at A B and C D. 
These outlines constitute a front elevation of the lower 
end of the molded panel, or the view as seen from a 
point exactly in front of the face of the raking bracket 
when finished and in its proper or final position. The 
outline or shape of the upper end of the panel would 
appear as a simple straight line in this view because it 
miters against a surface which is horizontal from front 
to back. AECDFE shows the entire front view of 
the molded panel. This view furnishes the means for 
the next step, which is to obtain a view at right angles 
to the face G H, and at the same at right angles to the 
lines of the rake 1ST 0. To do this, first continue the 
lines from the normal profile of panel in their vertical 
course till they intersect the upper line of the panel 
B F. These lines are omitted through the face of the 
bracket, the points only being indicated on the line E 
F. From the points thus established in E F, and from 
the points derived in the outlines A B and C D, carry 
lines at right angles to the raking cornice, producing 
them indefinitely, as shown. At right angles to the 
raking cornice, at any convenient place, draw the line 
H 1 and G 1 , setting off on it spaces corresponding to 
those established in H G, already described. Through 
the points in H' G 1 draw lines at right angles to it to 
the left, producing them until they intersect lines al- 
ready drawn from the outlines A B and C D and the 
points in the line E F. Through the points of inter- 
section thus obtained, as indicated by 1 7 in the lower 
left hand corner, 8 14 in the lower right hand corner, 8, 
9, 10, etc., in the upper right hand corner, and 1, 2, 3, 
4, etc. , in the upper left hand corner, trace lines, thus 
completing a view of the panel piece at right angles to 
its face. The next step to be taken is to develop a 
true profile of this panel, or in other words, a section 
at right angles to its lines, from which to obtain a 
stretchout for the required pattern. To do this, first 
assume any line, as P 0, at right angles to the lines of 
the view just obtained as the surface of the panel in 
the new profile. Upon this line extended, as at K, 
draw a duplicate of normal profile so that the points 1 
and S shall lie in it. Divide the profile K into the 
same number of spaces as in previous instances, and 
from these points carry lines through the face view in- 
tersecting them with lines of corresponding number, as 



Pattern Problems. 



19T 




Fig. 3S5.—The Pattern for a Raised Panel on the Face of a Raking Bracket 



198 



Tlie New Metal Worker Pattern Book. 



shown at L P and Q R. Then LPQE will be the 
true profile of the moldings along the face of the raking 
bracket. The student will observe that only half the 
profile is shown at K, as both halves are alike, one- 
half will answer all purposes if it be kept in mind while 
making the intersections by number that the points 
1-7 in one profile are 14-8 in the other. At any con- 
venient place lay off the stretchout of the true profile, 



as shown to the left by the line L M. Through the 
points in this line draw the usual measuring lines, as 
shown. Then, with the blade of the T-square placed 
parallel with the stretchout line and brought against 
the several points of intersection at the corners of the 
" View at Right Angles to the Face," cut correspond- 
ing measuring lines. Lines traced through the points 
thus obtained will produce the pattern shape, as shown, 



PROBLEM 90. 

The Patterns for a Diagonal Bracket Under Cornice of a Hipped Roof. 



In Fig. 



3S6 is shown a constructive section of the 
cornice of a hipped roof, under which the bracket L 
fits against the planceer and over the bed molding C. 
Fig. 387 shows an inverted plan of the angle of such a 
cornice, including two normal brackets B and C, and 
the diagonal bracket D, of which the patterns are re- 
quired. At A, in line with one arm of the cornice in 
plan, is also shown a duplicate of the profile of the 
normal bracket. E F represents the miter line of the 
planceer over which the diagonal bracket is required 
to fit. 

Two distinct operations are necessary in obtaining 
the patterns of the bracket D, one for the face pieces 
and the other for the sides. As the bracket is placed 
exactly over (or more properly speaking under) the 
miter in the cornice, one-half its width must be drawn 
on either side of the miter line, as shown in Fig. 387. 
Each half of its face thus becomes a continuation of 
the moldings forming the faces of the course of normal 
brackets of which it is a part. Therefore the normal 
profile X 8 of the bracket A is the profile to be used, 
and I Gr and J F form the miter lines for one half 
the face. 

The usual method in obtaining the pattern for the 
face piece would be to divide the profile of A into any 
convenient number of spaces and lay off a stretchout 
of the same upon any line drawn at right angles to the 
direction of the mold — that is, at right angles to I J or 
G F — after which lines should be dropped from the 
profile upon the miter lines and thence into the stretch- 
out. However, as the miter is a square miter, the 
short method is available ; hence the stretchout line is 
drawn at right angles to the horizontal line of the ele- 
vation X X, as shown at H Gr. The usual measuring 
lines are drawn and intei'sected with lines from points 
of corresponding number on the profile. Lines traced 



through the points of intersection, as shown by K M 
and L N, will give the pattern for half the face. 

The operation of obtaining or "raking" the pat- 
tern of the side is exactly similar to that employed in 
Problem 88, with the difference that while in Problem 
88 the side is elongated vertically, in the present in- 
stance (the cornice remaining horizontal, and the bracket 
being placed obliquely) it is elongated laterally or 
horizontally. The operation is also complicated by 
the addition of a profile at the back edge of the bracket 







Fig. S86.— Sectional View of the Cornice of a Hipped Roof, 
Showing Bracket. 

where it is required to fit over the bed molding of the 
cornice. To obtain the pattern of the side it is first 
necessary to ascertain the correct horizontal distances 
between the various points of the profile. The points 
already made use of in obtaining the face may be used 
for this purpose. Therefore, drop lines from each of 
these points vertically, intersecting the side of the 
bracket, or, what is the same thing, the center line E F, 
as shown in the plan, Fig. 387, by 1', 2', 3', etc. The 



Pattern Problems. 



199 



profile at the back of the bracket in the elevation must 
also be divided into a convenient number of spaces, as 
shown by the small figures, which must also be dropped 
upon E F, as shown, and numbered correspondingly. 



transfer the points and spaces from E F. Now from 
each point in the line E' F', erect lines vertically, in- 
tersecting lines of corresponding number previously 
drawn to the right from the elevation. Thus, lines 
drawn upward from the intersections 1, 2, 3, 4, etc., 
on the line E' F' intersect with horizontal lines 1, 2, 3, 
4, etc., while lines drawn upward from the intersec- 
tions 1', 2', 3', 4', etc., on the line E' F' intersect with 
horizontal lines l',2', 3', 4', etc. Lines traced through 
the points of intersection, as shown by R S P, will 
be the required pattern of the side. 

If it be desirable to ascertain the exact ancle to 
which to bend the edges or flanges of the bracket to 
fit against the planceerit may be accomplished in the 
following manner: Extend P of the pa.ttern of the 
side till it intersects the line from X of the side eleva- 




Firj, SS7. — Inxertcd Plan of Cornice and Method of Obtaining Patterns. 



From each of the points in the profile of the elevation 
carry lines indefinitely to the right, as shown. At any 
convenient point at the right of the plan, draw another 
plan of the diagonal bracket, so placed that its sides 
shall be parallel with the horizontal line X X of the 
elevation, all as shown, and upon its center line E' F' 



tion, as shown at Y. Upon the solid line X X in 
diagonal elevation establish any point, as T. Through 
the point T and at right angles to Y F draw a line in- 
tersecting the line Y P at U. 

As the angle of the plan shown in Fig. 387 is a 
right angle, construct a right angle, ABC, Fig. 388, 



200 



Tlie New Metal Worker Pattern Booh. 



and bisect it, obtaining the miter line B L. Now take 
the distance from Y to T m diagonal elevation, and 
place it on the miter line B L in Fig. 388, from B to 
D. At rignt angles to B L draw a line through the 
point D, intersecting the sides of the right angle A B 




Fig. S88.— Diagram for Obtaining Angles for Bending the Flanges. 

C at E and F. Now take the distance T U in diag- 
onal elevation and set it off from D toward B, locating 
the point H. Connect the points E, H and F ; then 
will the profile E H F in Fig. 388 represent a section 
across the hip at right angle to its rake and will also 
be the angle to be used in putting the straight parts of 



the face together, as shown Dy E' H' F' in Fig. 389. 
The angle which the sides of the bracket make with 
the planceer will be the complement of the angle H E 
D of Fig. 38S and may be obtained as follows : Paral- 
lel to B L, in Fig. 388, and through the point E, draw 




Fig. SS9. — Perspective View of Finished Bracket. 

I K, representing the vertical side of the bracket; then 
will the angle J E I represent the profile required for 
bending the flanges on the side of the bracket, the pro- 
file being shown in position by I' E' J' in Fig. 389. 

In Fig. 389 is shown a perspective view of the 
finished bracket as seen from below. 



PROBLEM 91. 

To Obtain the Profile of a Horizontal Return, at the Foot of a Gable, Necessary to Miter at Right 
Angles in Plan With an Inclined Molding: of Normal Profile, and the Miter Patterns of Both. 



In the elevation B C E D, and plan FGHKLI, 
of Fig. 390 is presented a set of conditions which 
necessitate a change of profile in either the horizontal 
or raking molding, in order to accomplish a miter 
joint at I H in the plan. In other words, the condi- 
tions are such that with a given profile, as shown by 
A', in the raking molding, the profile of the horizontal 
molding forming the return will require to be modi- 
fied, as shown by the profile A 2 , in order to form a 
miter upon the line I H in the plan. 

The reason for this is easily found. If a vertical 
line be erected from point 9 in profile A 2 it will be 
seen that each line emanating from a point in the nor- 
mal profile A 1 becomes depressed after passing this 
vertical line, more or less, according as its distance 
away from this line increases, all in proportion to the 
amount of rake or incline of the face molding, as 
shown by the dotted lines. If, on the contrary, the 
profile A 2 be considered as the normal profile, the pro- 



file A 1 will have to be changed or "raked," in thi3 
case increased in hight, in proportion to the inclina- 
tion. (These conditions are treated in the succeeding 
problem.) The vertical hight of the profile of the re- 
turn may be measured in the side elevation and com- 
pared with that of the inclined molding by measuring 
across the latter at right angles to the line B C. 

In this problem it is assumed that the profile as 
well as the pitch, or rake, of the cornice B are 
established and that the profile of the horizontal re- 
turn is to be modified, or "raked," to suit it. To 
obtain this profile, first draw the normal profile in. the 
raking cornice, as shown by A 1 , placing it to corre- 
spond to the lines of the cornice, as shown. Draw 
another profile corresponding to it in all parts, directly 
above or below the foot of the raking cornice, in line 
with the face of the new profile to be constructed, 
placing this profile A so that its vertical lines shall cor- 
respond with the vertical lines of the horizontal cor- 



Pattern Problems. 



201 



nice. Divide the profiles A and A 1 into the same 
number of parts, aud through the points thus obtained 
draw lines, those from A 1 being 
parallel to the lines of the raking 
cornice, and those from A intersecting 
them vertically. Through these points 
of intersection of like numbers trace a 
line, which gives the modified profile, 
as shown by A 2 . Then A J is the 
profile of the horizontal return, indi- 
cated by G II I F in the plan. It is also 
the elevation of the miter line I H of the 
plan. Therefore at any convenient point 
at right angles to the lines of the raking 
cornice lay off the stretchout M N of 
the i^rofile A 1 , through the points in 
which draw measuring lines in the 
usual manner. Place the T-square at 
right angles to the lines of the 
raking cornice, and, bringing it suc- 
cessively against the points in the 
profile A 2 , cut the corresponding measuring lines 
just described. Through the points of intersection 
trace a line, as shown by OPE. Then OPE 
will be the shape of the lower end of the raking 
cornice mitering against the return. For the pattern 
of the return proceed as follows : Construct a side 
elevation of the return, as shown by S V IT T, mak- 
ing the profile V U the same as the profile A" of the 
elevation. Let the length of the return correspond to 
the return, as shown in the plan by F I. In the pro- 
file V U set off points corresponding to the points in 
the profile A 3 as shown from B to D. At right angles 
to the elevation of the return lay off a stretchout of 
V IT, or, what is the same, of the profile A 3 , as shown 
by W X, through the points in which draw measuring 
lines in the usual manner. Placing the T-square 
parallel to this stretchout line, and bringing it success- 



ively against the points in V IT, cut tne corresponding 
measuring lines. Then a line traced through these 




Fig. $90.— To Obtain the Profile of a Horizontal Return at the 
Foot of a Gable, Necessary to Miter at Right Angles in Plan unth 
an Inclined Molding of Normal Profile, and the Patterns of Both. 

points of intersection, as usual, from Y to Z, will be 
the pattern of the horizontal return. 



PROBLEM 92. 



To Obtain the Profile of an Inclined Molding: Necessary to Miter at Right Angles in Plan with a Given 

Horizontal Return, and the Miter Patterns of Both. 



The conditions shown in this problem are similar 
to those in the one just demonstrated. In this, how- 
ever, the normal profile is given to the horizontal re- 
turn, and the profile or the raking cornice is modified 



to correspond with it. To obtain the new profile pro- 
ceed as follows: Divide the normal profile A", Fig. 
391, into any convenient number of parts in the usual 
manner, and from these points carry lines parallel to 



202 






The New Metal Worker Pattern Book. 



the lines of the raking cornice indefinitely. At any 
convenient point outside of the raking cornice, and at 



sq uare 




number of spaces. With the T-square at right angles 
to the lines of the raking cornice, and brought success- 
ively against the several points in this profile, cut 
corresponding lines drawn through the cornice from 
the profile A 1 . Then a line traced through these 
points of intersection, as shown by A 3 , will be the 
profile of the raking cornice. For the pattern of the 
foot of the raking cornice mitering against the return, 
take the stretchout of the profile A 3 and lay it off on 
any line at right angles to the raking cornice, as shown 
by P 0. Through the points in this stretchout line 
draw the usual measuring lines, as shown. With the 



Fig. 391— To Obtain the Profile of an Inclined Mold- 
ing Necessary to Miter at Right Angles in Plan 
with a Given Horizontal Return, and the Pat- 
terns of Both. 
H K J 

right angles to its lines, construct a duplicate of the 
normal profile, as shown by A 3 , which divide into like 



at right angles to the lines of the raking cor- 
nice, or parallel to the stretchout line, bring 
it successively against the points in the pro- 
file A 1 , which is also an elevation of the 
miter, and cut the measuring lines drawn 
through the stretchout P O. Then a line 
traced through the points of intersection, as 
shown by B' R 1 , will be the miter pattern 
of the foot of the raking cornice. 

For the pattern of the return proceed 
as follows : Construct an elevation of the 
return, as shown by F 1 G 1 K 1 H 1 , in di- 
mensions making it correspond to F G K 11 
of the plan. Space the profile A of the re- 
turn in the same manner as A 1 . At right angles to 
the lines in the return cornice draw any straight line, 
as M N on which lay off its stretchout, through the 
points in which draw measuring lines in the usual 
manner. Place the T-square at right angles to the 
lines of the return cornice, and, bringing it successively 
against the points in the profile A, cut the correspond- 
ing measuring lines. Through the points of intersec- 
tion trace a line, as shown by G 3 K\ In like manner 
draw a line corresponding to F 1 H 1 of the side eleva- 
tion. Then F 3 G 3 K 3 H 3 will be the pattern of the 
horizontal return to miter with the raking cornice, as 
described. 



PROBLEM 93. 

To Obtain the Profile of the Horizontal Return at the Top of a Broken Pediment Necessary to Miter 
with a Given Inclined Molding:, and the Patterns of Both. 



In Fig. 



392, C B D represents a portion of the 
elevation of what is known as a " broken pediment, ' ' 
the normal profile of whose cornice is shown at A 1 . 
With these conditions existing it becomes necessary to 



obtain new profiles for the returns at both the top and 
the foot. The method of raking the return at the foot 
has been described in'Problern 91, and the method of 
raking the return at the top is exactly the same. If, 



Pattern Problems. 



203 



in the designing of the pediment, the normal profile 
should oe placed in the return at the foot, as is some- 
times necessary, then the profile of the inclined mold- 
ing must be first obtained, which in turn must be con- 
sidered as a normal profile and used as a basis of ob- 
taining the third profile, that of the return at the top. 

In Fig. 392, let A 1 be considered as the normal 
profile of the inclined molding. Divide A 1 into any 
convenient number of parts in the usual manner, and 
through these points draw lines parallel to the lines of 
the cornice indefinitely. At any convenient point out- 
side of the cornice, and in a vertical line with the point 
at which the new profile is to be constructed, draw a 
duplicate of the profile of the raking cornice, as shown 
by A, which space into the same number of parts as 
A 1 , already described. From the points in A draw 
lines vertically, intersecting lines drawn from A 1 . Then 
a line traced through these several points of intersec- 
tion, as shown by A", will constitute the profile of the 
horizontal return at the top and also the miter line as 
-shown in elevation. If the normal profile were in the 
horizontal return at the foot of the pediment and the 
modified profile in the position of A 1 , 
it would be immaterial whether the 
normal profile or a duplicate of the 
modified profile were in the place of 
A by which to obtain the intersecting 
lines, as the projection of the points 
only is to be considered in this opera- 
tion, and that is the same in both cases. 
For the pattern of the inclined 
molding proceed as follows : At right 
angles to the lines of the raking 
cornice lay off a stretchout of the profile of the raking 
cornice A 1 , as shown by F G, through the points in 
which draw measuring lines in the usual manner. 
Place the T-square at right angles to the lines of the 
raking cornice, and, bringing the blade successively 
against the points in the profile A 2 , which is the miter 
line in the elevation, cut the corresponding measuring 
lines, and through these points of intersection trace a 
line, as shown by G H. Then G- H will be the pat- 
tern- of the top end of the raking cornice to miter 
against the horizontal return. For the pattern of the 
horizontal return the usual method would be to con- 
struct an elevation of it in a manner similar to that 
described for the return at the foot of the gable in the 
preceding demonstrations ; the equivalent of this, how- 
ever, can be done in a way to save a considerable por- 
tion of the labor. 



As the view of the miter line is the same in both 
the front and the side elevation the pattern may be de- 
veloped from the front just obtained in the following 
manner, with the result, however, that the pattern will 
be reversed : Draw the line K M perpendicular to the 




Fig. 392. — To Obtain the Profile of the Horrizontal Return on the 
Top of a Broken Pediment Necessary to Miter with a Given 
Inclined Molding, and the Patterns of Both. 



lines of the horizontal return, as it would be if shown 
in elevation. Upon K M lay off a stretchout of the 
profile A ! , all as shown by the small figures, and 
through the points draw the usual measuring lines. 
With the "["-square parallel to the stretchout line K M 



204 



The New Metal Worker Pattern Book. 



bring the blade successively against the points in the 
profile A', cutting the corresponding measuring lines. 
Through these points of intersection trace a line, as 



shown by N L, which will be the pattern of the end 
of the horizontal return to miter against the gable 
cornice, as shown. 



PROBLEM 94. 

To Obtain the Profile and Patterns of the Returns at the Top and Foot of a Segmental Broken Pediment. 



The preceding three problems treat of the various 
miters involved in the construction of angular pedi- 
ments. In Fig. 393 is shown an elevation of a curved 
or segmental broken pediment in which the normal 
profile is placed in the horizontal return at the foot. 
The profiles for the curved molding and for the return 
at the top can both be obtained at one operation in 
the following manner: Divide the normal profile 
ABC into any convenient number of parts, and from 
the points thus obtained draw lines at right angles to 
the horizontal line C F of elevation, as shown. At 
any convenient point draw G H, at right angles to 
A G, cutting them. With Q, the point from which 
the curve of the molding was struck, as center, strike 
arcs from the points in A B C, extending them in the 
direction of D indefinitely. From any convenient 
point in the arc A D, as L, draw a line to the center Q. 
From L draw L M, at right angles to L Q, upon which, 
beginning at L, set off the distances contained in H G, 
as shown by the small figures in L M. From the 
points of intersection where arcs struck from Q cut 
L Q draw lines at right angles to L Q. From the 
points in L M, and at right angles to it, drop lines 
cutting those of similar number drawn at right angles 
to L Q. A line traced through these points of inter- 
section, as shown by M K, will be the profile of 
curved molding. It will be observed that the points 
for obtaining the profile are where the perpendiculars 
dropped from L M intersect the lines drawn at right- 
angles to L Q, and not where the perpendiculars 
dropped from L M intersect the arcs. 

For the profile D E draw N D, parallel to J, or 
at right angles to N 0, and, starting from D, set off 
on D N the same points as are in G H. Drop perpen- 
diculars from these points to the arcs of similar num- 
bers drawn from A B, when a line traced through the 
points of intersection will form the desired profile, as 
show by D E. The normal profile is also drawn above 
G II and NDatX and Z to show that the same result 
is obtained by using the points in G H to set off on 
L M and N D as would be obtained by dropping the 
points from the profiles. The patterns for the returns 



would be obtained as described in the previous prob- 
lems. 




Fig. S9S.—To Obtain the Profiles and Patterns of the 
Returns at the Top and Foot of a Segmental Broken 
Pediment. 



Problems describing the method of obtaining the 
pattern for the blank for the curved molding will be 
found in Section 2 of this chapter. 



Pattern Problems. 



PROBLEM 95. 

From the Profile of a Given Horizontal Molding-, to Obtain the Profile of an Inclined Molding: Necessary 
to Miter with it at an Octagon Angle in Plan, and the Patterns for Both Arms of the Miter. 



Another example wherein is required a change of 
profile in order to produce a miter between the parts 
is shown in Fie;. 394. In this case the ansrle shown 




S94.— From the Profile of a Given Horizontal 
Molding to Obtain the Profile of an Inclined 
Molding Necessary to Miter with it at an Octa- 
gon Angle in Plan, and the Patterns for Both 
Arms of the Miter. 



in plan between the abutting members is that of an 
octagon, as indicated by B C D. To produce the 
modified profile and to describe the patterns proceed 
as follows: In the side B C draw the normal profile A, 



as indicated, and in the corresponding side, as shown in 
elevation by N L K, draw a duplicate profile, as shown 
by A 1 . Divide both of these profiles into the same num- 
ber of parts, and from the points in each 
carry lines parallel to the lines of mold- 
ing in the respective views, producing 
the lines drawn from profile A until 
they meet the miter line C X. From 
the points thus obtained in C X erect 
lines vertically until they meet those 
drawn from profile A 1 , intersecting as 
shown from to L. Through these 
points of intersection draw the line 
O L, which will be the miter line in 
elevation corresponding to C X of the 
plan. From the points in L carry 
lines parallel with the raking molding 
in the direction of P indefinitely. At 
any convenient point outside of the 
raking cornice draw a duplicate of the 
normal profile, as shown by A 2 , placing 
its vertical line at right angles to the 
lines of the raking cornice. Divide the 
profile A" into the same number of 
spaces as employed in A and A', and 
from these points carry lines at right 
angles to the lines of the raking cornice, 
intersecting those of corresponding 
number drawn from the points in L. 
Trace a line through these intersections, 
as shown from B to S. Then E S 
will be the required profile of a raking 
cornice to miter against a level cornice 
of the profile A at an angle indicated by 
BCD in the plan, or an octagon angle: 
Forthe pattern of. the level cornice, 
at right angles to the arm B C in the 
plan lay off a stretchout of the profile 
A, as shown by E F, through the 
points in which draw the usual measur- 
ing line. With the T-square at right 
angles to B C, bringing the blade suc- 
cessively against the several points in 
X C, cut corresponding measuring lines drawn through 
E F. Then a line traced through these points, as shown 
from H to G, will be the required pattern of the hori- 
zontal cornice. In like manner, for the pattern of 



206 



Tlie New Metal Worker Pattern Booh. 



the raking cornice, at right angles to its lines lay off a 
stretchout of the profile E S, as shown by U T, 
through the points in which draw measuring lines in 
the usual manner. With the T-square at right angles 
to the lines of the raking cornice, and brought success- 



ively against the points in the miter line L, as shown 
in elevation, cut the corresponding measuring lines. 
Then a line traced through the points thus obtained, 
as shown by W V, will be the required pattern for the 
rakins; cornice. 



PROBLEM 96. 

From the Profile of a Given Inclined Molding:, to Establish the Profile of a Horizontal Molding to Mite; 
with it at an Octagon Angle in Plan, and the Patterns for Both Arms. 



In Fig. 



395, let BCD be the 
which the two moldings are to join, 
in elevation, and A or A 1 the nor- 
mal profile of the raking mol 
To form a miter between mokliin 
meeting; under these conditions 



angJ 




Fig. S95.—From the Profile of a Given Inclined Mold, 
to Establish (he Profile of a Horizontal Molding 
to Miter with it at an Octagon Angle in Plan, and 
the Patterns for Both Arms. 



change of profile is required, 
profile for the horizontal arm 



To obtain the modified 
and the miter line in 



elevation proceed as follows : Draw 
the normal profile A with its vertical 
side parallel to the lines in the plan of 
the arm E X D C, corresponding to the 
front of the elevation. Draw a dupli- 
cate of the normal profile in correct 
position in the elevation, as shown 
by A 1 . Divide both of these profiles 
into the same number of parts, and 
through the points in each draw lines 
parallel with the plan and with" the 
elevation respectively, all as indicated 
by the dotted lines. From the points 
in the miter line of the plan C E, 
obtained by the lines drawn from the 
profile A, carry lines vertically, inter- 
secting the lines drawn from A 1 . 
Then a line traced through the inter- 
sections thus obtained, as shown from 
N to 0, will be the miter line in ele- 
vation. From the points in 1ST 
carry lines horizontally along the arm 
of the horizontal molding N U Y, 
as shown. At any convenient point 
outside of this arm, either above or 
below it, draw a duplicate of the nor- 
mal profile, as shown by A 2 , which 
divide into the same number of parts 
as before, and from the points carry 
lines vertically intersecting the lines 
drawn from N 0, just •described 
Then a line traced through these 
points of intersection, as shown by 
T S, will give the required modified 
profile. 

For the patterns of the arm Y 1ST 
U proceed as follows: At right 
angles to the same, as shown in plan 
by "W E C B, lay off on any straight line, as G F, a 
stretchout of the profile T S, all as shown by the small 



Pattern Problems. 



207 



figures V, 2 3 , 3 ! , etc. Through these points draw 
measuring lines in the usual manner. With the T-square 
parallel to the stretchout line, and brought against the 
points of the miter line E C in plan, cut corresponding 
measuring lines, as indicated by the dotted lines, and 
through these points of intersection trace a line, as 
shown by K H. Then K H will be the shape of the 
end of Y N TJ to miter against the raking molding. 

It will be easily understood that the points as 
found upon the line E C are just the same as would be 
obtained there if the newly obtained profile were drawn 
into the plan of the arm C B W E and the points were 



dropped from it to the line E C according to the rule. 
For the pattern of the raking molding, at right angles 
to the arm N Z V O in the elevation lay out a stretch- 
out, L M, from the profile A'. Through the points 
in this stretchout draw measuring lines in the usual 
manner. Place the T-square parallel to the stretchout 
line, and, bringing it against the several points in the 
miter line in elevation N 0, cut corresponding measur- 
ing lines, as indicated by the dotted lines. Then a 
line traced through these points of intersection, as 
shown by P E, will be the shape of the cut on the arm 
N Z V to miter against the horizontal molding. 



PROBLEM 97. 

The Miter Between the Moldings of Adjacent Gables of Different Pitches upon a Pinnacle with Rectangular Shaft. 



The problem presented in Figs. 396 and 397 is 
one occasionally arising in pinnacle work. The figures 
represent the side and end elevations of a pinnacle which 




Fig. S96. — Side Elevation of Rectangular Pinnacle, Showing the 
Miter Between the Moldings of Adjacent Gables. 

is rectangular, but not square. All of its faces are 
finished with gables whose moldings miter with each 



other at the corners, and which are of the same bight 
in the line of their ridges, as indicated by L M and L 1 
M 1 . Whatever profile is given to the molding in one 
face of such a structure, the profile of the gable in the 
adjacent face will require some modification in order to 
form a miter. In Fig. 396 let A be the normal profile 
of the molding placed in the gable of the side elevation. 
Before the miter patterns can be developed it will first 
be necessary to obtain the miter line or joint between 
the moldings of the adjacent gables as it will appear 
in the elevation, to accomplish which proceed as fol- 
lows : Draw a duplicate of A, placing it in a vertical 
position directly below or above the point at which the 
two moldings are to meet, as shown by A 1 . Divide 
both of these profiles into the same number of parts, as 
indicated by the small figures, and through these points 
draw lines intersecting in the points from H to K, as 
shown. Then a line traced through these intersections 
will be the miter line in elevation. For the pattern of 
the molding of the side gable lay off at right angles to 
H M a stretchout of the profile A, as shown by B C, 
through the points of which draw the usual measuring 
lines. Place the T-square at right angles to the lines 
of the molding, or, what is the same, parallel to the 
stretchout line, and, bringing it against the several 
points in the miter line H K, cut corresponding meas- 
uring lines. Then a line traced through these points, as 
shown by D E, will be the shape of the cut at the foot 
of the side gable to miter against the adjacent gable. 

The next step is to obtain the correct profile 
of the molding on the adjacent gable. H K having 
been established as the correct elevation of the miter, its 



208 



Hie New Metal Worker Pattern Book. 



outline may now be transferred, with its points, to the 
end elevation of the pinnacle, as shown at H' K 1 , Fig. 
397, reversing it, because it appears here at the right 
side of the gable, whereas it appeared at the left of the 
other. Draw a duplicate of the normal profile, as shown 
at A\ placing its vertical lines at right angles to the 
lines of the gable, and divide it into the same spaces as 
in the first operation. From these points draw lines at 
rio-ht angles across the molding, which intersect with 

DO ^ 

lines drawn parallel to the molding from the points in 
the miter line H 1 K 1 . Then a line traced through 
these points of intersection will form the required modi- 
fied profile, as shown by W X. 

For the pattern of the molding of the end gable 
proceed as follows : At right angles to the lines of the 
raking cornice lay off a stretchout of the profile "W X, 
as shown by P E, through the points in which draw 
measuring lines in the usual manner. With the 
T-square at right angles to the lines of the raking 
cornice, bringing it successively against the points in 
K' H 1 , cut corresponding measuring lines. Then aline 
traced through these points of intersection, as shown 
from S tp T, will be the pattern required. 




Fig. 397.— End Elevation of Rectangular Pinnacle, Showing Same 
Miter as in Fig. 396. 



PROBLEM 98. 

The Miter Between the Moldings of Adjacent Gables of Different Pitches upon an Octagon Pinnacle. 



This problem differs from the preceding one in 
that the angle of the plan is octagonal instead of 
square, but like it requires a change of profile in one 
of the gables in order to effect a miter. In Figs. 398 
and 399 are shown a quarter plan of pinnacle and the 
elevations of two adjacent gables of different widths 
but of similar bights. Let A 1 B 1 F 1 G 1 D' of Fig. 
398 be a correct elevation and A B C G be a quarter 
plan of the structure. In that portion of the plan cor- 
responding to the part of the elevation shown to the 
front draw the normal profile E, placing its vertical 
side parallel to the lines of the plan. Divide it into 
any convenient number of spaces, and through these 
points draw lines parallel to the lines of the plan, cut- 
ting C O 1 , the miter line in plan, as shown. In like 
manner place a duplicate of the normal profile, as shown 
by E 1 in the elevation. Divide it into the same num- 
ber of equal parts, and through the points draw lines 



parallel to the lines of the raking cornice, which pro- 
duce in the direction of X indefinitely. Bring 
the T-square against the points in C O 1 , and with it 
erect vertical lines, cutting the lines drawn from E 1 , as 
shown from X to 0. Then a line, X 0, traced through 
these points of intersection will be the miter line in 
elevation. 

For the pattern of the miter at the foot of the 
wide gable or gable shown in elevation proceed as fol- 
lows : At right angles to the lines of the gable cornice 
lay off a stretchout of the profile E 1 , as shown by H K, 
through the points in which draw the usual measuring 
lines. Placing the T-square at right angles to the lines 
of the cornice, or, what is the same, parallel to the 
stretchout line, and bringing it against the several 
points in X 0, cut corresponding measuring lines. 
Then a line traced through the points of intersection 
thus obtained, as shown from L to M, will be the pat- 



Pattern Problems. 



209 



tern of tne miter at the foot of the gable shown in ele- 
vation. For the modified profile of the gable molding 




Fig. 398.— Quarter Plan and Elevation of Octagon Pinnacle, Show- 
ing Miter Between Moldings of Adjacent Gables of Different 
Pitches. 



upon the narrow side proceed as follows : Draw a cor- 
rect elevation of the narrow side, reproducing therein 



the miter line 1ST from Fig. 398 (reversing tne same), 
as shown by R P in Fig. 399, and through the points, 
also reproduced from N" 0, cany lines parallel to the 
lines of the gable cornice indefinitely, as shown. Draw 
a duplicate of the normal profile at any convenient 
point outside of the gable cornice, as shown by E% 
placing its vertical side at right angles to A 2 R, or the 
lines of the cornice. Divide E 2 into the same number 
of parts as used in the other profiles, and through the 
points draw lines at right angles to the lines of the cor- 
nice, intersecting the lines drawn from P R. Through 




Fig. 399. — Elevation of Narroiv Side of Octagon Pinnacle, Showing 
Same Miter as in Fig. 398. 



these points trace a line, as indicated by E 3 , which will 
be the modified profile. 

To lay out the pattern take the stretchout of 
E 3 and lay it off on any straight line drawn at right 
angles to the lines of the cornice, as S T, and through 
the points in it draw the usual measuring lines. Place 
the T-square at right angles to the lines of the gable 
cornice, and, bringing it against the points in P R, cut 
the measuring lines, as indicated by the dotted lines. 
Then a line traced through these points of intersection, 
as shown by IT T, will be the pattern for the molding 
at the foot of the gable on narrow side. 



210 



Tlie New Metal Worker Pattern Book. 



PROBLEM 99. 

The Patterns for a Cold Air Box in which the Inclined Portion Joins the Level Portion Obliquely in Plan. 



The conditions of the problem are clearly shown 
in the plan and side elevation of Fig. 400, in which Z 
B C is the elevation and X C D' Y is the plan of the 
level portion of a cold air passage joining a furnace 
just above the floor line. The inclined portion of the 
air passage or box is required to join the level portion 
at the angle Z A E of the side elevation, and at the 
angle Y A' E' when viewed in plan. These conditions 
are in many respects similar to those given in Problem 
95, with the difference, however, that in this case the 
joint or miter between the level and the inclined por- 
tions does not appear as a straight line in the plan. It 
may be here remarked that the solution of this prob- 
lem is more a matter of drawing than of pattern cutting, 
as nothing can be more simple than the cutting of a 
miter between two pieces of rectangular pipe when the 
required angle between them is known. This problem 
is capable of two solutions, both of which will be 
given, leaving the reader to choose which is the more 
adaptable to his requirements. 

First Solution. — As above intimated, before the 
pattern can be developed it will be necessary to 'make 
careful drawings, in the preparation of which a knowl- 
edge of the principles of orthographic projection is 
necessary. (See Chapter III). 

To proceed, then, with the drawings, first draw a 
plan and elevation of as much of the furnace as is 
necessary to show its connection with the cold air box, 
placing each part of the plan directly under its corre- 
sponding part in the elevation, so that as soon as any 
new point is determined in either of the views its posi- 
tion can be located in the other by means of a perpen- 
dicular line dropped from one view to the other. Upon 
the plan set off the width of the box b and draw parallel 
lines from the side of the furnace body to the right 
indefinitely, and upon the elevation set off its hight, a, 
from the floor line up, and draw A Z. A vertical line 
from the point X of the plan will give the point Z upon 
the elevation, or, in other words, show how far the 
curve of the furnace body cuts into the top and 
bottom surfaces of the cold air box. Next, upon the 
elevation locate the point A the required distance from 
the side of the body according to specification and find 
its position in the plan by means of a vertical line, as 
shown. From the point A in both views lines must 
be drawn to represent the angle or deflection of the 



pipe as it would appear in those views. Thus the ele- 
vation would show the slant, which is determined by 
the two dimensions c and d. Therefore from the point 
A of the elevation erect a perpendicular line equal to 
the required hight e, from the top of which draw a 
horizontal line to the right of a length equal to the 
amount of slant d, thus locating the point E, which 
connect by a straight line with A. Then will A E 
represent the angle of the inclined portion of the pipe 
as it appears in the elevation. But according to the 
requirements the pipe is also to have an offset a dis- 
tance equal to e — that is, the point E of the elevation 
is nearer the observer than the point A. Therefore 
from A' of the plan draw a line forward the amount of 
the offset, from the end of which draw a line to the 
right, in length equal to d, or in other words till it 
comes directly under the point E of the elevation, thus 
locating that point in the plan, and draw A' E', which 
will show the apparent angle in the plan. 

The depth and width of the oblique portion of the 
box will next demand attention. At right angles to 
the line A E of the elevation set off the depth of the 
box a and draw a line to represent the lower near cor- 
ner of the box, which continue downward until it cuts 
the floor line, as shown at D ; then draw A D, which 
represents the miter cut for the side of the box. At 
right angles to A' E' of the plan set off the width /;, as 
shown, and draw a line parallel to A' E' intersecting the 
line from X at B', as shown, and draw A' B', which 
gives the plan of the miter cut across the top of the 
box. As the point D of the elevation is in the same 
vertical plane as A it may now be dropped into the 
plan, intersecting with the line showing the front side 
of the box in that view, as shown at D' ; and the point 
B' of the plan, being on a level with A', may be pro- 
jected into the elevation, where it would intersect with 
the line showing the top of the box at B. A line 
drawn from D' of the plan parallel to A' E' (shown 
dotted) will then show the position of the lower near 
angle of the inclined portion of the box, and a line 
from B of the elevation parallel to A E will show the 
position in that view of the further top corner of the 
box. 

The position in the two views of the remaining 
angle of the inclined portion of the box may be ascer- 
tained in several ways : The width b may be set off 



Pattern Proocems. 



211 



from T)' of the plan and a line drawn which will inter- 
sect with X B' continued, as shown at C ; thence it 
may be projected into the elevation at C, as shown; 
or the width a may be set off from B of the elevation, 
thus locating the line which intersects with the floor at 
C, which point may be dropped into the plan, thus 
locating the point C ; or, again, B C may be drawn 
parallel to A D, or D' C may be drawn parallel to A' 
B', all producing the same result. 

In the case in Problem 95, above referred to, it 
was noted that if the normal profile is adhered to in the 
level arm, the profile of the gable mold must be changed 
or "raked" before a perfect miter joint can be ob- 
tained. What is true in the case of the gable miter is 
equally true in the case of the furnace pipe — a correct 
profile or cross section of the box must be developed 
in order that a correct stretchout may be obtained for 
use in cutting the miter of the inclined arm of the pipe. 
As neither the plan nor the elevation, which have been 
correctly obtained, gives the true length of the inclined 
piece — that is, the true distance from A to E — it will 
be necessary to obtain still another elevation, in which 
such distance is correctly shown. As A' E' of the 
plan gives the horizontal distance between the 
points A and E, and c represents the vertical 
distance between them, if a right angled triangle 
be constructed with A' 
E' as a base and the 
bight c as the perpendic- 
ular, itshypothenuse will 
then give the desired 
measurement. Such a 
triangle properly forms 
part of an oblique eleva- 
tion which may be pro- 
jected from the plan in 
the following manner : 
Parallel to A' E', at any 
convenient distance 
away, draw a line to rep- 
resent the level of the 
floor, as shown; above 
which, at a distance equal 
to a, draw another paral- 
lel line, X 2 A 2 , represent- 
ing the hight of the hori- 
zontal arm of the pipe. 
Above the line X 2 A 2 , at 
a hight equal to c, draw 
still another line, upon 
which the point E is sub- 
sequently to be located. 




Fig. 400.- 



-Plan and Elevations of a Cold Air Box in. Which the Inclined Pcrtion Joins the 
Portion Obliquely in Plan. — First Solution. 



Level 



212 



Tlie New Metal Worker Pattern Booh. 



Now drop lines from all the points of the plan at right 
angles to A' E', intersecting each with its corresponding 
line of the new elevation, thus locating each point of the 
miter in that view As points D' and C are upon the 
floor, their position will be found at D 2 and C 2 . Like- 
wise lines from A' and B' will locate those points in 
the upper surface of the horizontal pipe, as shown at 
A 2 and B 2 , where they are also shown to be in the side 
elevation. A line dropped from E' will also locate that 
point at its proper hight, as shown at E 2 . A line con- 
necting A 2 and E 2 will then be the hypothenuse above 
alluded to and be the correct length sought. As all 
edges or corners of the pipe are necessarily parallel, 
lines drawn from B 2 C 2 and D 2 parallel to A 2 E 2 will 
complete this part of the elevation as far as necessary. 
In these, as in all geometrical drawings, lines showing 
parts concealed from view by other parts are always 
shown dotted. Lines from X and Y locate those points 
in the new elevation and show that, while a correct 
elevation of the inclined arm of the pipe has been ob- 
tained, the view of the horizontal portion is oblique, the 
space between X 2 and Y 2 showing the open end to fit 
against the furnace body. 

Having now obtained a correct oblique elevation, 
the next step is to obtain a correct profile upon any 
line, as F H, drawn at right angles across the pipe, 
which may be accomplished in the following manner: 
From each point upon the line of the section F, G, J 
and H project lines' parallel with the direction of the 
pipe to a convenient point outside the elevation, as 
shown at the left, across which draw a line, x y, at 
right angles to them as a base from which to measure 
distances from front to back. 

Assuming its crossing with the line from G (point 
1) to represent the near angle of the pipe, set off from 
x on the line from F the horizontal breadth of the pipe 
b, thus locating point 4, which corresponds to the 
point F in the elevation. In like manner on the line 
from H set off from y the distance o of the plan, locat- 
ing the point 2, which corresponds to point H of eleva- 
tion, and draw the lines 1 4 and 1 2. The distance 
of point 3 from line x y is equal to distance b plus the 
distance o, or in other words, draw the line 2 3 par- 
allel to 1 4 and the line 4 3 parallel to 1 2, thus 
locating the point 3. 

Having now a profile and a correct elevation 
of the miter, nothing remains but to lay off a 
stretchout, as shown, upon the line H K and drop 
the points in the usual manner from the profile to 
the miter line A a B 2 C 2 D 2 , thence into the measuring 



lines of the stretchout, all as clearly shown in the 
drawing. 

As the plan shows all the dimensions of the hori- 
zontal arm of the pipe, the pattern for that can be de- 
veloped in the usual manner. To avoid confusion a 
duplicate of that part of the plan has been transferred 
to Fig. 401, where a stretchout of the normal profile 
is laid off at right angles to the lines of the pipe, into 
which the points are dropped from the miter line A B 
C D. In the normal profile of course the distances 1 
4 and 2 3 are equal to b and the distances 1 2 and 4 
3 equal to a of Fig. 400. 




D 12 

PLAN 
Fig. 401.— Plan and Patte7ii of Level Arm of Cold Air Box. 

It may be noted here that, as is the case in all 
raked profiles, the dimensions and shape of the profile 
obtained from the oblique elevation differ somewhat 
from those of the normal profile shown in Fig. 401, 
and that their stretchouts are therefore necessarily dif- 
ferent. 

Second Solution.— It may be asked naturally, is 
there no way of producing a miter without a change of 
profile, just as a carpenter would saw off the ends of 
two square sticks of timber of the same section and 
produce a perfect miter at an oblique angle ? There 
is, but the method of doing it is not so apparent as the 



Pattern Problems. 



213 



one just described. To accomplish this a drawing or 
view must be obtained, in which the surface of the 
paper represents a plane common to both arms of the 



shown in Fig. 402, in which the plan shown in Fig. 
400 has been reproduced, but turned around in such a 
manner as to facilitate the projection from it of an end 




Fig. 402.— Patterns of Cold Air Box. —Second Solution. 

pipe. As three points determine the position of a 
plane, it will be seen at once that such a plane passes 
through the points Z, A and E of the side elevation, 
Fig. 400. The best means of obtaining this view is 



elevation, all of which is clearly shown in the drawing. 
This view shows the offset e and the rise c of the 
oblique portion of the pipe. The new view, which 
will give the required conditions, is obtained by look- 
ing at the pipe in a direction at right angles to A E of 
the end elevation, and is obtained as follows : Parallel 
to A E at any convenient distance away draw A' E', 
which make equal to A E by means of the lines drawn 
at right angles to A E, as shown. Upon the line E' E 
set off from E' the slant d as given in the side eleva- 
tion and plan, Fig. 400, locating the point E 2 , and 
draw the line E 2 A'. From all points of the profile or 
end view of the horizontal pipe, 1. 2, 3 and 4, project 
lines also at right angles to A E, continuing them 
across the line A' E', and make A' Y 2 equal to A Y 
of the plan. Then A' Y 2 will be the length of the 
horizontal arm in the new view and A' E 2 will be the 
length of the inclined arm, both lying in the same 
plane, and the angle E 2 A' Y 2 will be the angle at which 
the two arms meet. Under the above conditions, then, 
a line which bisects that angle, as A' C, will be the 



214 



Tlte New Metal Worker Pattern Book. 



miter line between the two arms. As the two arms of 
the miter are symmetrical, the view can be completed, 
if desired, by drawing lines parallel with A' E 2 from 
the points of intersection with the lines from the end 
view with the miter line A' C. As 1 2 3 4 is the 
profile from which the short arm was projected in the 
new view, a stretchout may now be taken from it and 



laid off on any line at right angles to C W and the points 
dropped in the usual manner, all as shown. If desired, 
the stretchout may also be laid off at right angles to 
the inclined arm and the pattern for this piece thus 
developed from the same miter line, although the miter 
cut A B D A is the same in both pieces, one simply 
being the reverse of the other. 



PROBLEM ioo. 

The Patterns for the Inclined Portion of a Cold Air Box to Meet the Horizontal Portion Obliquely in Plan. 



This problem is here introduced on account of 
the similarity of its conditions with those of the one 
immediately preceding, although, as its patterns are 
obtained entirely without the use of profiles, it does 
not properly belong in this connection. Its solution 
will serve to show what widely different 
means may be employed to obtain the 
same ends. In the preceding case the 
miter cut was obtained without refer- 
ence to the miter at the upper end of 
the oblique arm. In this case the 
oblique portion is required to join, at 
its upper end, with another arm like 
and exactly parallel with the arm join- 
ing its lower termination. 

Under such conditions it follows 
that the planes of the upper and lower 
miters must be parallel, and, therefore, 
that miter cut at the upper end of 
either of the faces of the oblique por- 
tion must be parallel with that at the 
lower end of the same. Advantage 
may be taken of these conditions to 
obtain a very simple solution of the 
problem, as will be seen below. 

The first requisite is, of course, a 
correctly drawn elevation and plan in 
which all the points in each are duly 
projected from corresponding points in 
the other view. In Fig. 403 is shown 
a plan and elevation of the box, with 
the lines of projection connecting cor- 
responding points in each, all of which may be con- 
structed very much as described in the preceding 
problem. The inclined arm is required to have a rise 
equal to a of the elevation and a forward projection 
equal to b of the plan. Corresponding points in the 



two views are lettered alike. Thus the elevation shows 
clearly that it is an elevation of the front A B F E of 
the plan, with the back CDHG dotted behind, while 
the plan shows clearly A B D C of the elevation with 
the bottom EFH6 dotted below. 



ELEVATION 
A 



SECTION 




Fig. JflS.- 



Patterns for the Inclined Portion of a Cold 
Portion Obliquely in Plan. 



Air Box to Meet the Level 



The first important information to be derived from 
the correctly drawn views is that the front and back 
are the same, likewise the top and bottom are alike. 
The patterns of the top and front are given separately, 
upon the supposition that joints will be made at all of 



Pattern Problems. 



215 



the angles; should they be wanted in one piece they 
could readily be connected. As all the surfaces of the 
inclined portion of the pipe are oblique to the given 
view, only some of their dimensions will be correct as 
they appear on the paper. An inspection of both 
elevation and plan will show that the lines A C and 
B D are both horizontal and parallel, and, therefore, 
correct as they appear in the plan, and may be used 
as given in the construction of a pattern of the top 
piece. The shortest distance between these two lines 
will be represented by a line at right angles to both, 
as M N. Since the point N in the line B D is higher 
than the point M of the line A G, by the distance a 
of the elevation, it will be necessary to construct the 
diagram J L K in order to get the correct distance be- 
tween the points M and 1ST. J K is made equal to the 
distance M N, as indicated by the dotted lines. K L 
is equal to the rise given in the elevation; hence the 
distance J L represents the true distance between the 
points M and 1ST. Upon the continuation of the line 
M N of the plan set off the distance J L, as shown at 
J' L'. Through each of these points lines are drawn 
parallel to A C and B D of the plan. The line A' C 
is made equal to A C, and B' D' is made equal to B D 



by means of the dotted lines drawn parallel to M N. 
This pattern is completed by connecting the point A' 
with B' and C with D'. 

In developing the pattern of the side A B F E 
the same course might be pursued, beginning with the 
lines A E and B F, whose lengths are correctly given 
in the elevation, but for the sake of diversity another 
method has been employed. Beginning with the 
known fact that the point B is higher than the 
point A, as shown by a in the elevation, con- 
struct a diagram, P R, making P equal to 
and parallel with A B of the plan, and R equa 
to a, thus giving R P as the correct length of th 
line represented by A B of the plan. From the 
points E and F draw, at right angles to E F, the lines 
E S and F T indefinitely. Since the distances A E 
and B F are the same and are correctly given in the 
elevation, take that distance between the feet of the 
dividers, and placing one foot at the point R describe a 
small arc, cutting the line E S in the point S. By 
repeating this operation from the point P, the point T 
is established in the line F and T. Lines connecting 
the points R S, S T and T P will complete the pat- 
tern of the front and back. 



PROBLEM ioi. 

The Pattern of a Hip Molding: upon a Right Angle in a Mansard Roof, Mitering Against the Planceer 

of a Deck Cornice. 



Let Z X Y Y in Fig. 404 be the elevation of a 
deck cornice, against the planceer of which a hip mold- 
ing, shown in elevation by U W Y T, is required to 
miter. Let the angle of the roof be a right angle, as 
shown by the plan Q ~D A 1 , Fig. 405, D N represent- 
ing the plan of the angle over which the hip molding 
is to be placed. This angle is also shown by B A of 
the elevation. As the only view which will show the 
correct angle at which the hip molding meets the plan- 
ceer is a view at right angles to the line D 1ST, the first 
step in the development of the patterns will be to con- 
struct such a diagonal elevation. Assume any point, 
as A, in the elevation on any line representing a plain 
surface in the profile of the roof, as B A. Through A 
draw a horizontal line indefinitely, as shown by L A C. 



From B, the point in which the line A B meets the 
planceer, drop a vertical line, cutting the horizontal 
line drawn through A at the point C, all as shown by 
B C. Produce the line of planceer W Y, as shown by 
W Y'. Draw a duplicate of the plan, Q D A 1 in Fig. 
405, in such a manner that the diagonal line D N shall 
lie parallel to the horizontal line drawn through A, al' 
as shown by Q 1 D 1 A 3 . At right angles to the line D 
A 2 , at any convenient point, as A 2 , draw the line A 3 
C, in length equal to the distance A C in elevation, 
and through C draw a line parallel to D 1 A 2 , as shown 
by I ISP, cutting the diagonal line D 1 ISP in the point 
1ST 1 . Then D 1 ISP represents the diagonal plan of that 
part of the hip from B to A in the elevation. From 
ISP erect a perpendicular, ISP M, which produce until it 



216 



Tlie New Metal Worker Pattern Book. 



meets the line carried horizontally from the planceer 
in the point B'. In like manner from D 1 erect a per- 
pendicular, which produce until it meets the horizontal 
line L C in the point L. Connect L and B 1 , as shown, 
which will constitute the desired oblique projection 
of A B. 

The next step will be to construct a section of the 
hip molding upon a line at right angles with it, as G 



the side D' A', from which erect a line perpendicular 
to D' N l , as shown by E F, which produce until it 
meets the horizontal line L C in the point L', and 
thence carry it upward parallel to L B 1 , cutting G H 
in the point F\ On either side of F 1 lay off a space 
equal to F E of the diagonal plan, as shown by F 1 E 1 
and F' E'. Through these points E 1 and E 3 draw lines 
to K, the intersection of the lines L B 1 and G H. From 




Fig. 404. — The Pattern of a Hip Molding in a Mansard Roof, Mitering Against the Planceer of a Deck Cornice. 



H, assumed at any convenient point. It might be 
supposed that in such a section the two fascias of the 
hip molding would be at right angles to each other, as 
they undoubtedly would appear in the plan Q D A 1 of 
Fig. 405 or in a section on any horizontal line, as L M. 
The object of this part of the demonstration is to show 
exactly what that angle would be and how to obtain it. 
Assume any point in the diagonal plan, as E, in 



K as a center describe the curve of the roll of the re- 
quired diameter. Upon the lines K E 1 and K E 2 set 
off from K a distance sufficient to make the desired 
width of fascia, thus completing the profile of the hip 
molding in the diagonal elevation. 

Space one-half of this profile, as G E', in the usual 
manner, through the points in which carry lines par- 
allel to L B 1 , cutting the line of planceer W Y', which 



Pattern Problems. 



217 



is the miter line of the roll. The edges of the fascia 
will of course miter with the lower edge of the fascia 




Fig. 405. — Plan of the Fascias and Angle of the Mansard Shown 
in Fig. 404. 

at the top of the mansard, shown in profile at B E 3 , all 
as shown by the dotted lines projected from E\ At 



right angles to the line L B 1 draw the straight line S R, 
upon which lay off a stretchout of the profile in the 
usual manner, and through the points draw measuring 
lines. "With the T-square parallel to this stretchout 
line, or, what is the same, at right angles to the lines 
of the molding in the diagonal elevation, and, bringing 
it successively against the points in W Y 1 , cut corre- 
sponding measuring lines drawn through the stretchout. 
The measuring lines 7 and 8 are cut from the inter- 
section of the fascia of the hip with lines projected 
from E 3 as above explained. Then a line traced 
thi*ough these points, as shown in the engraving, 
as shown by J 1 P J, will be the pattern of 
the hip molding mitering against the horizontal 
planceer. 



PROBLEM 102. 

The Pattern for a Hip Molding upon a Right Angle in a Mansard Roof, Mitering Against a Bed Molding 

at the Top. 



Let A C B, in Fig. 406, be the section of a por- 
tion of a mansard roof, the elevation of which is shown 
to the left, and let P E be any bed molding whose 
profile does not correspond to or member with the 
molding used to cover the hips, a section of the hip 
molding being shown at Z Y C. 

The solution of this problem will be accomplished 
by means of a "true face" of the roof, rather than 
by means of a diagonal elevation as in the problem 
immediately preceding this. Therefore, supposing 
the section A C B to give the correct pitch of the 
roof, the first step will be to obtain the true face, or 
elevation of the roof as it would appear if tipped or 
swung into a vertical position, for the purpose of get- 
ting the correct angle at A 1 B 1 F 1 . 

To do this reproduce the section of mansard and 
bed molding as a whole at a convenient point below, 
but so turned as to bring the faces of the roof into a 
vertical position, maintaining the same distance be- 
tween the points A and B as shown by A 2 and B\ 
Project lines horizontally to the left from this section 
for the true face, marking the lines from the points 
A 2 andB 2 . From A of the original section carry a line 
across intersecting the line A' B 1 at the point A 1 . 
Next drop line from A 1 and B 1 vertically intersecting 
lines of corresponding letter, as shown by the dotted 



lines. Then A" B 9 F' will be the correct angle upon 
which to construct the corner piece and develop the 
miter line between the hip molding and the bed mold- 
ing of the deck cornice. 

The next step will be to obtain a correct section 
of the hip molding upon a line at right angles to the 
line of the hip. To do this it is necessary to first 
construct a diagonal section through the hip. At any 
convenient place lay off a plan of the angle of the roof, 
as shown by D 1 F D 2 in Fig. 407, and through this angle 
draw a plan of the hip, as shown by F K. From D 1 erect 
a line perpendicular to F D', as D' C 2 , in length equal 
to D C of the section. Through C 2 , parallel to D 1 F, 
draw C 2 K, producing it until it cuts the line repre- 
senting the plan of the hip. From the points F and K 
in the lines representing the plan of the hip erect per- 
pendiculars, as shown by F L and K C 3 . Draw L s 
parallel to F K, as shown at the base line of a diag- 
onal section. From C 3 erect a perpendicular, C 3 E', in 
length equal to C E of the original section. Connect 
E 1 L. Then L C 3 E 1 will be a diagonal section of a 
portion of the roof, and L E 1 will be the length of the 
hip through that portion. At right angles to L E 1 
draw M H 1 , upon which to construct a correct section 
of the hip molding. Take any point, as G in the line 
F D 1 , at convenience, and from it erect a perpendicular 



218 



The New Metal Worker Pattern Book. 



to F K, cutting F K in the point H, and produce it 

also until it cuts the base line of the diagonal section 
L C 3 , as shown, and from this point carry it parallel to 
the line L E', representing the pitch of the hip, until it 



will be obtained by which the angle contained between 
the facias of the hip molding may be determined. 
Therefore from H 1 on either side set off the distance 
H G of the plan, as shown by G 1 and G\ Through 




Fig. 406.— The Pattern of a Sip Molding Upon a Right Angle in a Mansard Roof, Mitering Against a Bed Molding. 



crosses the line M H 1 , cutting it in the point II'. 
Since D 1 F D a represents the angle in plan over which 
the hip molding is to fit, and since H G is the meas- 
urement across that angle, if the distance H G be set 
off from H 1 either way in the diagonal section, points 



these points draw lines representing the fascias of the 
hip molding, as shown by O G 1 and G\ Add the 
fillets and draw the roll according to given dimensions, 
all as shown. 

In the true face, Fig. 406, draw a half section of 



Pattern Problems. 



219 



the hip molding as derived from Fig. 407, as shown. 
M 3 H 3 corresponds to M H' of the diagonal section. 




Fig. 407. — Method of Obtaining Correct Gross Section of Hip in 
Fig. 406. 



Space this profile into any convenient number of parts 
in the usual manner, and through the points draw lines 
parallel to the lines of the hip molding indefinitely. 



Place a corresponding portion of the profile of the hip 
molding in the vertical section, as shown, in which 
M' H 3 also corresponds to H 1 M in the diagonal section. 
Divide this section into the same number of equal 
parts, and through the points draw lines upward until 
they intersect with the profile of the bed molding, as 
shown between P 3 and B\ From the points in P 3 B 3 carry 
lines horizontally, intersecting the lines drawn from 
the profile in the true face. Then a line traced 
through these points of intersection will be the miter 
line between the hip molding and the bed molding, as 
seen near B 3 in elevation. 

For the pattern proceed as follows: At right 
angles to the line of the hip molding, as shown in the 
true face, lay off a stretchout of the hip molding, as 
shown by S R, through the points in which draw the 
usual measuring lines. Place the T-square at right 
angles to the lines of the hip molding, and, bringing 
it successively against the several points in the miter 
line, as shown in elevation, cut corresponding measur- 
ing lines, which will give the pattern for the roll and 
fillets, as shown from TJ to V. In like manner place 
the T-square against the point X in the true face, 
which is the point of junction between the flange of 
the hip molding and the apron of the bed molding 
corresponding to points 9 and 10 of the profile, and 
cut the corresponding measuring lines. The pattern 
is then completed by drawing a line from W to V and 
T toU. 



PROBLEM 103. 

Patterns for the Top and Bottom of the Hip Bar in a Skylight. 



In the upper part of Fig. 408 is shown the trans- 
verse section of a skylight in which A B represents a 
portion of the ventilator or finish at the top, and C D 
the curb or finish at the bottom. The section also 
shows the side elevation of a " common " bar whose 
profile is at F. The plan immediately below shows a 
corner of the skylight with one of the hip bars, H K, 
the patterns for which are required. It will be neces- 
sary first to see that the plan is correctly projected 
from the elevation, and afterward that a diagonal ele- 
vation of the hip bar be obtained from this plan, be- 
fore the correct or raked profile of the hip bar can be 
obtained. 

Draw a duplicate of normal profile F with its cen- 
ter line on the center line of the hip, as shown at F 1 , 



as a means of obtaining the lateral projection of all its 
points, numbering corresponding points in both profiles 
the same. Number the intersections of all the points 
in the normal profile F with the top and bottom of the 
skylight finish, as shown by the small figures in A B 
and C G. From each of the points in the profile F 1 
carry lines parallel to the center line of the hip in either 
direction, intersecting lines of corresponding number 
dropped vertically from both the miters of the trans- 
verse section to the plan. Lines traced through these 
points of intersection will give the miter lines at top 
and bottom as they appear in plan. 

At right angles to the lines of the hip carry lines, 
as shown, by means of which to construct the diagonal 
elevation. Assume any line, as E 1 G 1 , as the base 01 



220 



Tlie Neio Metal Worker Pattern Booh. 



horizontal line of the diagonal elevation representing 
E G of the section. At E 1 erect a perpendicular upon 



the horizontal molding at the top whose profile is 
shown at A B, it will be found most convenient to 




Fig. 408. — Plan and Section of a Skylight and Patterns for the Hip Bar. 



which to obtain the hights of the various points in the 
upper miter. As the hip bar is required to miter with 



carry all the points of the upper profile to the vertical 
line A B, as shown by 1, 2, 3 1 , 4, 5 and 6', and after- 



Pattern Problems. 



221 



ward to transfer them, as shown, to the line A 1 B', 
keeping the perpendicular hight from E 1 to B 1 equal 
to E B. From all the points in A 1 B 1 carry lines hori- 
zontally — that is, parallel to E 1 G' — to the right indefi- 
nitely, as shown. These lines will then represent a 
partial elevation of the top molding A B in the 
diagonal elevation. Lines from each of the points in 
the plan of the upper miter at H may now be carried 
parallel to H E 1 until they intersect with lines of cor- 
responding number drawn from A 1 B 1 . Lines connect- 
ing the points of intersection will give the required 
miter line at the top of the hip bar. 

From each of the points obtained in this miter 
line carry lines parallel to B 1 G', or the rake of the hip 
bar, and intersect them with lines projected parallel to 
II E 1 from the lower miter in plan at K. Lines con- 
necting these points of intersection will give the re- 
quired miter line at the bottom of the hip bar. 

It now remains only to obtain the correct profile 
of the hip bar before a stretchout can be obtained. 
To accomplish this, draw any line cutting the lines of 
the hip bar in the diagonal elevation at right angles, as 
shown at R. Upon this line, and above or below the 
hip bar, as shown at F 2 , draw a duplicate of the normal 
profile F, from the points in which carry lines at right 
angles to the hip bar, cutting lines of corresponding 
number in the same. Then lines connecting the points 
of intersection will give the raked profile, as shown at R. 

On account of limited space the important details 



in Fig. 408 are necessarily small, but great care has 
been taken in the preparation of the drawing, and all 
the points in the several views of both miters have 
been carefully numbered, so that the reader will have 
no difficulty in following out the various intersections 
from start to finish. The profile and the two miter 
lines now being in readiness, the pattern may be de- 
veloped in the usual manner, as follows : Upon any 
line drawn at right angles to the hip bar, as L M, lay 
off a stretchout of the profile R, as shown by the small 
figures, through which draw the measuring lines. 
Keeping the blade of the -square parallel with L M, 
bring it successively against the points of intersection 
previously obtained in the upper and lower miters and 
cut corresponding measuring lines. Then lines traced 
through the various points of intersection, as shown by 
N and P Q, will constitute the required patterns. 

It may be noticed that while most of the points 
from the normal profile F come squarely against the 
inner beveled surface of the curb G, the points 1 and 
2, representing the vertical portion of the bar, pass 
over the curl) to a point bej^ond. The line from point 
2, therefore, intersects at both 2 and 2 1 , which points 
are duly carried through the views of this miter at K 
and G' and finally into the pattern, as shown ; from 
which it may be seen that the miter pattern may be 
cut as shown by the solid line from P to Q, or that 
portion from point 2 to 3 may be cut as shown by the 
dotted line. 



PROBLEM 104. 



Pattern for the Top of a Jack Bar in a Skylight. 



The jack bar in a skylight is the same as the 
bar in respect to its profile, and the miter 
at its lower end with the curb. At its upper end, 
however, it is required to miter against the side of the 
hip bar instead of against the upper finish of the sky- 
light. As the hip bar occupies an oblique position 
with reference to the jack bar, it is evident that a 
perfect miter between the two could not be effected 
without a modification or raking of the profile of the 
hip bar, all of which has been demonstrated in the 
preceding problem. 

It may be here remarked that the raking of the 
profile of the hip bar is done not so much to affect a 
perfect joint with the top finish as to make a perfect 



miter with the jack bar, or what is the same thing, 
that the surfaces indicated by 2 3 of the profile of the . 
hip bar in Fig. 408 shall lie in the same plane with 
that portion of the profile of the jack bar. However, 
as the raked hip bar presents exactly the same appear- 
ance when viewed in plan as a bar of normal profile, 
it will not be really necessary, so far as the miter cut 
on the jack bar is concerned, to perform the raking 
operation. 

In Fig. 409 is shown a sectional and a plan view 
of a portion of a skylight containing the miter above 
referred to. The normal profile of the jack bar shown 
at F and F 1 is not exactly the same in its proportions 
as that of the preceding problem, but possesses the 



222 



Tlie New Metal Worker Pattern Booh. 



same general features. 



The view of the bar given in 



the section from A to B represents an oblique eleva- 
tion of that side of the hip bar which is toward the 
jack bar. From B to D the view shows the side of the 
jack bar, while beyond D is shown a continuation of 
the full hip bar with its profile correctly placed in po- 
sition at F 3 . 

The first step before the pattern can be laid out 
is to obtain a correct intersection of the points in the 
plan, as at B', and afterward an elevation of the same, 
as shown at B. Draw a normal profile of the jack bar 
in correct position in the plan, as shown at F 1 . Also 
place a profile of the hip bar in the plan of the same, 
as shown at F 2 . As only the lateral projection of the 
points are here made use of a normal profile will 
answer as well as the raked profile shown, as above 
intimated. Number all the points in both profiles 
correspondingly, and from the points in each carry 
lines respectively parallel to their plans, intersecting 
as shown at B 1 . From the points of intersection of 
like numbers erect lines vertically into the sectional 
view, cutting lines of corresponding number drawn 
from the points in the profile F parallel to the lines of 
the rake, as shown near B. It will be seen that both 
sides of the profile F 1 intersect with one side of the 
profile F 2 , both sets of intersection being numbered 
alike, as l 1 , 2 1 , 3', etc. This gives rise to two miter 
lines at B in the sectional view. The line correspond- 
ing to the intersections on the upper side of the jack 
bar are here numbered l 2 , 2 2 , 3 2 , etc., while those 
points belonging exclusively to the lower intersection 
are numbered 3 3 , 5 s and 6 s . 

A stretchout of the normal profile F may now be 
laid off on any line, as G H, drawn at right angles to 
the elevation of the jack bar, through which the usual 
measuring lines are drawn. Now place the blade of 
the T-square parallel to G H, aud, bringing it against 
the various points in the two miter lines above de- 
scribed, cut corresponding measuring lines, carrying the 
points from the upper miter line into one side of the 
pattern and those from the lower one into the other 
side ; then lines connecting the points of intersection, 
as shown from K to L, will constitute the required 
miter cut. 

As it is desirable to cut the miter on the jack bar 
so as to fit over the hip bar (that is, so as not to cut the 
hip bar at all) and m order to prevent the surface from 
4 to 5 of the jack bar from lapping on to a like por- 
tion of the hip bar, as shown between the points 4 1 , 5 1 
and x in the plan, the line from point 4 of F 1 is al- 



lowed to intersect with the line from 5 of F 2 , as shown 
at x, which point is carried into the sectional view and 
thence into the pattern, where it intersects with lines 4, 
as shown by x, so that the cut in the pattern is from 
a; to 5 instead of from 4 to 5. For the same reason, 




/ '-Ju-U 
j //hi V 

Ml/,' //K 

J ' 
j 




Fig. 409. — Section and Plan of Miter at the Top of the Jack Bar in 
a Skylight and Pattern of the Same. 



if it is desired to prevent the surfaces 2 3 from over- 
lapping the line from 2 of F 1 may be intersected 
with 3 from F 2 , as shown at y, and carried into the 
pattern, as shown, producing the cuts in the pattern 
shown by the dotted lines 3 y in the place of those 
shown from 3 to 2. 



Pattern Problems. 



223 



PROBLEM 105. 

The Pattern of a Hip Mold Upon an Octagon Angle in a Mansard Roof, Mitering Against a Bed 

Molding of Corresponding Profile. 



This problem, like many others pertaining to man- 
sard roofs, may reach the pattern cutter in drawings 
either more or less accurate, and in different stages of 
completion. Certain facts, however — viz. , the profiles 
of the moldings, the pitch of the roof and the angle in 
plan — must be known before the work can be accom- 
plished; but with these given the pattern cutter will 
have no difficulty in drawing such elevations as are 
necessary to produce the required patterns. 

In Fig. 410, let A B C D be the given section of 
the mansard trimming shown, A C the profile of 
the bed molding and apron, and B D E the pitch of 
the roof. According to the statement of the problem 
above the angle of the plan is octagonal ; it might be 
a special angle either greater or less than that of an 
octagon, but the principle involved and the operation 
of cutting the patterns would be the same. As in all 
other problems connected with mansard trimmings, the 
first requisite is an elevation of the " true face," in 
order to obtain the correct angle between the bed mold- 
ing and the hip molding. A normal elevation, such 
as is likely to be met with in the architect's drawings, 
is shown in the engraving at the left of the section, 
merely for purposes of design. In obtaining the true 
face, shown below, it is best to use the section and plan 
only. Therefore, redraw the section as shown im- 
mediately below it, placing the line of the roof in a ver- 
tical position, all as shown. From all the points of this 
section lines may now be projected horizontally to the 
left, as the first step in developing the required true 
face. Immediately above the space allotted to the 
elevation draw a plan of the horizontal angle, as shown 
by I E 1 K. As it will be impracticable to include the 
entire profile of the roof in the drawings, some point 
must be assumed at a convenient distance below the 
bed mold, as D, from which to measure hight and pro- 
jection, which locate also in the section below, as shown 
at D 1 , making B' D 1 equal to B D. From A draw a 
line at right angles to the line of the roof, meeting it 
at B, which point may be assumed for convenience as 
the upper limit of that part of the roof under consider- 
ation. Now, from the point B drop a vertical line, 
which intersect with one drawn horizontally from D, as 
shown at E ; then D E will represent the projection. 



From the lines I E 1 and E 1 K, upon lines at right 
angles to each, set off the projection D E, as shown at 
I F and K H ; through the points F and H draw lines 
parallel to the first lines, meeting in G ; then a line from 
G to H will represent the plan of the angle or hip of 
that portion of the roof of which B D is the profile. 
Now, to complete the true face of that part of the 
roof drop a line from the point E 1 intersecting the line 
from B' at L, and one from G intersecting the one from 
D 1 at M ; then the angle B" L M will be the correct 
angle of the miter between the bed mold and the hip 
mold. 

As in Problems 101 and 102 preceding, it will next 
be necessary to obtain a correct section of the hip mold 
on a line at right angles to the line of the hip. To avoid 
confusion of lines, this operation is shown in i^o-. 411 
in which E 2 G 1 , the base line, is made equal to E 1 G of 
the plan in the previous figure. At the point E 2 erect 
a perpendicular, making it equal in hight to B E of the 
sectional view. Connect B 2 with G 1 , which will give 
the correct angle of the hip of the roof. As a means 
of constructing a correct section at right angles to this 
line, assume any two points on the original plan, as N 
and 0, equidistant, from G and connect them by a 
straight line, cutting the angle or hip line in P. Set off 
from G 1 on the line G 1 E 2 of Fig. 411 a distance equal 
to G P of the plan, as shown at P', from which draw a 
line parallel to the hip G 1 B\ Next intersect these 
two lines by another at right angles at any convenient 
point, as shown by P 2 Q. From the point P 2 set off the 
distances P 2 O 1 and P 2 N 1 , making them equal to P 
and P N. Connect the points O 1 and N 1 with E, which 
is the intersection of P 2 Q with the hip line ; then the 
angle 0' E N 1 will be a correct section of the roof upon 
the line P 2 Q or upon any line cutting the hip at rio-ht 
angles, upon which the finished profile of the hip mold 
may now be constructed as follows : Set off the pro- 
jections of the fascia and fillet as given in the sectional 
view, Fig. 410, from the lines E O 1 and E N 1 , contin- 
uing their lines to the center line P 2 Q. From the inter- 
section S as a center, with a radius of the bed mold 
describe the roll. 

As stipulated in the statement of this problem 
the profiles of the bed mold and hip mold are to cor- 



224 



Tlie New Metal Worker Pattern Book. 




tKq. 410.— The Pattern of a Hip Molding Upon an Octagon Angle, Mitering Against a Bed Molding of Corresponding Profile, 



Pattern Problems. 



225 



respond. By this it is understood that the curves of 
their molded surfaces are alike and struck with the same 
radius, and so placed as to member or miter. 

As the curve of the bed mold is only a quarter 
circle, while that of the hip mold is nearly three- 
quarters of a circle, it will be seen that the quarter 
circles in each half of the hip mold next adjacent to the 
fascias and fillets will miter with the arms of the bed 
mold on either side of the miter, and that a small space 
in the middle of the roll will remain between them, 
which must be mitered against the planceer, and the 
object of the operation shown in Fig. 411 is to deter- 
mine exactly what this space is. The dotted lines 
from S to the points 10 drawn at right angles 
to R ! and R X' show the limit of the quarter 
circles or the parts that must miter with the bed mold, 
while the space between them (10 to 10) shows the 
part that must miter against the planceer. It might be 
supposed that the angle between the fascias of the hip 
mold, to fit over the angle of a mansard which is octag- 
onal in plan, would be octagonal, but the demonstra- 
tion shows that while the angle N G of the plan, 
Fig. 410, is that of an octagon, the angle X 1 R O 1 , 
Fig. 411, is greater, because the distance N' O 1 is equal 
to X 0, while the distance R P 2 is less than G P, B P' 
being at right angles to the line of the hip and G 1 P' 
being oblique to it. 

The true face, Fig. 410, may now be completed, 
as follows : Upon any line, as S 1 T, drawn at 
right angles to L M, representing the face-of the roof, 
draw a duplicate of one-half the profile of the hip mold 
obtained in Fig. 411, placing the point S upon the line 
L M, as shown. Lines drawn through the angles of 
this profile parallel to L M will intersect with lines 
from corresponding points from the profile A' C, pre- 
viously drawn, giving the miter line J X and com- 
pleting the elevation of the true face. 



Upon any line at right angles to the line L M, as 
U V, lay off a stretchout of the complete hip mold as 
obtained from the half profile. S 1 T, through which 
draw measuring lines as usual. Drop lines from the 
points from 1 to 10 of the profile, parallel to L M, cut- 
ting the miter line; then, with the T-square placed at 
right angles to L M and brought successively against tho 
points in J X, intersect them with lines of corresponding 
number in the stretchout; then lines traced through the 
points of intersection, shown by cl b and a c, will give 
the pattern for that part of the profile from 1 up to the 
point 10. The pattern of that portion of the roll which 
miters against the planceer must; be obtained from the 
diagonal section of the hip. From points 10, 11 and 
12 in Fig. 411 carry lines parallel to G 1 B a intersecting 
the line of the planceer, as shown at ~W. It is only 
necessary to ascertain how much shorter the lines 11 
and 12 are than the line 10, and then to transfer these 
distances to the pattern. This can be done by drop- 
ping lines from the intersection of points 11 and 12 
with the planceer, in Fig. 411, at right angles to G 1 B% 
cutting line 10. These distances can then be trans- 
ferred to line 10 of the pattern, Fig. 410, measuring 
down from the point 10 of pattern already obtained,, 
after which they may be carried parallel to U V into 
the measuring lines 11 and 12, thus completing the 
pattern. 

This portion of the work is necessarily very minute 
in the drawing, but it will be easily seen, in applying 
the principle to other similar cases, that if the angle of 
the plan I E' K were less than that shown, for 
instance, if it were a right or an acute angle, a 
greater distance or more points would occur between 
the points 10 and 10, and further, that if the angle of 
the roof were less steep a greater curve or dip 
would occur between those points (a to b) of the 
pattern. 



PROBLEM 106. 

The Pattern of a Hip Molding: Upon an Octag-on Angle of a Mansard Roof, Mitering- Upon an Inclined 

Wash at the Bottom. 



In Fig. 412, let D B of the section represent the 
wash surmounting the base molding at the foot of a 
mansard roof, the inclination of the roof being shown 
by B A. The plan of the angle of the roof B 2 K B 2 , 
as specified, is that of an octagon, but so far as prin- 
ciple and method are concerned, it may be any angle 



whatever. The profile of the hip mold as giver, in the 
original drawings will most likely be drawn as fitting 
over an octagonal angle — that is, over the angle as given 
in the plan of the building. As explained in the 
problem preceding this, a section through the angle of 
the roof at right angles to the line of the hip must be 



226 



The New Metal Worker Pattern Book. 
.C 1 




G 1 ' H 2 D'\ 

Fig. 412.— The Pattern of a Hip Molding Upon an Octagon Angle, Mitering Upon an Inclined Wash at the Bottom. 



Pattern Problems. 



227 



obtained, to which the profile of the hip mold must be 
adjusted before going ahead. The difference between 
such a section and the angle in plan may seem trifling, 
but will be found to increase as the pitch of the roof 
decreases, and in a low hip roof will be found to be 
considerable. Hence the original detail of the hip 
mold must be accepted only so far as it gives width 
and depth of fascias and fillets, and diameter or radius 
of the roll, while the angle between the fascias must 
be adjusted to the true section across the hip as above 
stated. The method of doing this is shown in Fig. 
413, and the principles involved therein are explained 
in the previous problem in connection with Fig. 411, 
and need not, therefore, be repeated here. 

The first operation will consist in obtaining the 
"true face" of the roof in the usual manner, viz.: 
Assume any point upon the section of the roof, as A, 
at a convenient distance above the base, as a point 
from which to measure hight and projection. Eedraw 
the section of the roof immediately below the first one, 
placing it in a vertical position and locating thereon 
the point A, as shown by A'. From the points A", B 1 
and D 1 project lines horizontally to the left, thus ob- 
taining all the bights in the true face. It will be 
necessary next to complete the plan, to do which first 
obtain the projection of the points in the section upon 
any horizontal line, as the one drawn through B, which 
can be done by dropping vertical lines from the points 
A and D, cutting it as shown at I and C. Assuming 
the line B 2 K B 2 of the plan to represent- the point B 
of the section, set off upon any lines at right angles to 
the lines B 2 K these projections — that is, make B 2 I' 
equal to B I, and B 2 C 1 equal to B C. Through 
these points draw lines parallel to B 2 K, intersecting 
and forming the line P G, whicn is the plan of the 
angle over which the hip mold is required to fit. From 
the points P, K and G, which represent upon the angle 
of the roof the points A, B and D of the section, drop 
lines vertically into the true face intersecting the 



horizontal lines previously drawn from A 1 , B 1 and D', 
as shown; then P 1 K' T will be the correct angle at 
which to construct the miter of the half of the hip 
mold belonging to this face of the roof, and K 1 G" 
IF P will represent a corresponding elevation of the 
wash. 

The elevation of the true face may now be com- 
pleted by placing one-half the profile of the hip in 
correct position — that is, with its base line or fascia at 
right angles to the hip line P 1 K 1 , the point K coming 
on the line. Through the points Y, S and T project 
lines parallel to the hip line. To show the intersection 
of the hip mold with the wash, first place a duplicate 
of the half profile of hip mold in the sectional view, as 
shown by Y 1 R 1 T 1 ; then divide the curved portion of 
both profiles into the same number of equal spaces and 
number all the points correspondingly, as shown. 
From these points drop lines downward parallel with 
the lines of the respective views, those in the sectional 
view cutting the line of the wash B 1 B 1 . From these 
points of intersection carry lines horizontally, intersect- 
ing the lines dropped from the profile Y S T. Then a 
line traced through these points of intersection, as 
shown by Y 2 S 2 T 2 , will be the miter line formed by 
the junction of the hip molding with the wash. At 
right angles to the line of the hip molding in the true 
face lay off a complete stretchout of the hip molding, 
as shown by U V. Through the points in it draw 
measuring lines in the usual manner. Place the T- 
square parallel to this stretchout, or, what is the same, 
at right angles to the line of the hip molding, as shown 
in true face, and, bringing it successively against the 
points in the miter line Y" S 2 T 2 , cut the corresponding 
measuring lines. Then a line traced through these 
points of intersection, as shown from W to Z, will be 
the cut to fit the bottom of the hip molding. 

The normal elevation may be completed, if desired 
by means of projections from the plan and the section, 
as shown. 



PROBLEM 107. 

Pattern for a Hip Molding Mitering Against the Planceer of a Deck Cornice on a Mansard Roof 
Which is Square at the Eaves and Octagon at the Top. 



In Fig. 414 is shown the method of obtaining the 
miter against the planceer of a deck cornice formed by 
the molding covering a hip, which occurs between the 



main roof and that part which forms the transition 
from a square at the base to an octagon shape at the 
top. The roof is of the character sometimes employed 



228 



Tlie New Metal Worker Pattern Book. 



upon towers which are square in a portion of their hight 
and octagon in another portion, the transition from 
square to octagon occurring in the roof. The hip 
molding in question covers what may be called a transi- 
tion hip, being a diagonal line starting from one of the 
corners of the square part and ending at one of the 
corners of the octagon above. A carefully drawn plan, 
together with a section through one of the sides of the 
roof, giving the pitch, will be the first requisites to 
solving the problem, both of which are shown in the 
engraving. The first operation will be the construc- 
tion of a section upon the line of the hip, which may 
be done as follows : Assume any point, as A, in 
the section of the roof from which to measure 
hight and projection. If a horizontal line from A 
and a vertical line from the top of the roof surface 
B be intersected in C, then B C will represent 
the hight and A C the projection of the part of 
the roof assumed. Set off the projection C A at 
right angles to the top line of the plan C E, as shown 
by C A 1 , and carry a line through A 1 parallel to C E 
till it cuts the plan of hip E Q at D; then D E will 
form the base and B C the hight of the required sec- 
tion, which may be obtained for convenience by lines 
projected from D E at right angles, all as shown. The 
line B' D~ then represents the real angle at which 
the hip mold meets the planceer or level line at 
the top 

The next operation will consist in obtaining a cor- 
rect section of the hip mold from the data given and in 
placing it in correct position in the diagonal section. 
Take any point, G, in the plan at a convenient distance 
from the angle W D A 1 . Set off G 1 at the same dis- 
tance from the angle on the opposite side. From the 
points G and G 1 carry lines at right angles to and cut- 
ting D~ C in the points H 2 and 0'', and from these 
points carry them parallel with the line D~ B l indefi- 
nitely. At right angles to D 2 B' draw a line, as shown 
by Z H 1 , intersecting with the lines last drawn in the 
points H 1 and 0. From H 1 , along the line II 1 H a , set 
off a distance equal to PI G of the plan, and from O, 
in the line Z H 1 , set off a distance equal toO 1 G 1 of the 
plan, as shown by G 3 . Connect the intersection of 
Z IT and I) 2 B' with the points G 3 and G 2 , which will 
give the correct section through the angle of the roof. 
Having thus determined the angle of the hip molding 
finish, a representation of it is indicated in the drawing 



by adding the flanges and the roll. Since the miter 
required is the junction between the hip molding, the 
profile of which has just been drawn, and a horizontal 
planceer, the remaining step in the develojunent of the 
pattern consists simply in dividing the profile into any 
convenient number of parts, and carrying points against 
the line of the planceer, as shown near B', and thence 
carrying them across to the stretchout, as indicated. 
It is evident, however, upon inspection of the eleva- 
tion, that the apron or fascia strips in connection with 
the planceer which miter with the flanges of the hip 
molding will form a different joint upon the side cor- 
responding to the transition piece of the roof than 
upon the side corresponding to the normal pitch of the 
roof, owing to the difference in pitch of these two 
sides. To obtain the lines for this miter an additional 
section must be constructed, corresponding to a center 
line through the transition piece, as shown by W L in 
plan. Prolong C D 2 , as indicated, in the direction of 
W, and lay off W L', equal to W L of the plan. From 
L' erect a perpendicular, as shown by L' B', equal to 
C B of the original section. Connect W and B 2 , 
against the face of which draw a section of the apron or 
fascia strip belonging to the planceer, as shown, and from 
the points in it carry lines parallel to B 2 B' until they 
intersect lines drawn from the flange of the hip molding 
lying against that side of the roof, all as indicated by 
U X. From these points carry lines, cutting corre- 
sponding lines in the stretchout. The lines of the fascia 
behonging to' the other side are the same as if projected 
from the normal section at B, or as they appear in the 
elevation. Having obtained these points proceed as 
follows : At right angles to the lines of the molding in 
the diagonal section lay off the stretchout of the hip 
molding S T, and through the points draw the usual 
measuring lines, as shown. Place the T-square at right 
angles to the lines of the molding, or, what is the same, 
parallel to the stretchout line, and, bringing it success- 
ively against the points formed by the intersection of 
the lines drawn from the hip molding and the planceer 
lineB 1 , cut the corresponding measuring lines, as shown. 
In like manner bring the T-square against the points U 
and X, above described, and Y and V, points corre- 
sponding with the opposite side of the hip molding, 
and cut corresponding lines. Then a line traced 
through these several points of intersection, as shown 
by U 1 X' Y' V, will be the pattern sought. 



Pattern Problems. 



229 




Fig. 4U.— The Pattern for a Hip Molding Mitering Against the Planceer of a Deck Cornice on a Mansard Roof Which is Square at the 

Eaves and Octagon at the Top. 



230 



Tlie New Metal Worker Pattern Book. 



PROBLEM 108. 

Patterns for a Hip Molding: Mitering Against the Bed Molding- of a Deck Cornice on a Mansard Roof 

which is Square at the Base and Octag-onal at the Top. 



The problem presented in Fig. 415 is similar to 
that described in the previous problem, with the dif- 
ference that a bed molding is introduced in connection 
with the planceer against which the hip molding is to 
be mitered. MEM' represents a plan of the roof at 
the top, while LDM' represents a horizontal line at 
the point A of the section, assumed at convenience 
somewhere between the top and the bottom for the 
purpose of measurement. The intersection of the 
lines M L and E D prolonged would indicate the cor- 
ner of the building at the bottom of the roof, the 
structure being square at the base and octagonal at 
the top. 

The first step in the development of the pattern 
is to obtain a correct section of the roof on the line of 
one of the hips. Therefore, at any convenient point 
lay off E 3 D 3 of Fig. 416 equal to'D E of the plan. 
From the point E 3 erect a perpendicular, E 3 B\ in 
length equal to C B of the section of the roof. Con- 
nect B 2 and D 3 , which will be the pitch of the hip 
corresponding to the line D E of the plan. Since the 
section D s E 3 B 2 has been constructed away from and 
out of line with the plan, it will be necessary to re- 
produce a portion of the plan in immediate connection 
with the section, as shown by I 1 H A 3 C 2 . This can 
be done by tracing, or any means most convenient. 
From the point H in this plan lay off on either arm 
the points I and I', equally distant from it and con- 
veniently located for use in constructing the profile of 
the hip molding. From the points I and I 1 erect per- 
pendiculars to H C 2 , cutting it in the points K and 0, 
which prolong until they meet the base D 3 E 3 of the 
diagonal section, from which points carry them paral- 
lel to the inclined line D 3 B 2 indefinitely. At right 
angles to the inclined line D 3 B 2 draw a straight line, 
O 1 K 1 , cutting the lines last described in the points 
K 1 and 0'. From K 1 , measuring back on the line 
I s K 1 , set off the point F, making the distance from 
K 1 to F the same as from K to I of the plan. From 
l in the line I 1 F set off the distance O 1 F, equal to 
I 1 of the plan. From these points I 3 and F draw 
lines meeting the line 1 K 1 at the point of its inter- 



section with the line D 3 B 2 . Complete the profile of 
the hijD molding, as indicated, laying off the width of 
the fascias on these lines, adding the roll and edges. 

The next step in the development of the pattern 
is to draw a " true face " of the roof. In performing 
this operation it matters not whether the actual sur- 
face of the roof be used or the surface of the fascias. 
In this case the points A and B of Fig. 415, by which 
the depth and projections of the pitch are measured, 
are taken on the surface of the fascia. For the true 
face transfer the section A B to a vertical position, 
as indicated by A 2 B', Fig. 415, in connection with 
which the bed molding against which the hip mold- 
ing is to miter is also drawn, as shown. From the 
several points in this vertical section draw horizontal 
lines, which intersect by vertical lines dropped from cor- 
responding points in plan. Then D 2 E 2 X is the true 
face of that part of the roof corresponding to D E M' M 2 
of the plan. In connection with the vertical section just 
described, place a half profile of the hip molding, a 
true section of which has been obtained by the process 
already explained in Fig. 416, and also place a dupli- 
cate of this portion of the profile in connection with 
the true face. Space both of these profiles into the 
same number of parts, and from the several points 
in each cany lines upward parallel respectively to the 
lines of the views in which they appear; the lines 
from the profile in the vertical section cutting the bed 
molding, and the lines from the profile in the true face 
being continued indefinitely. From the points of in- 
tersection in the bed molding carry lines horizontally, 
intersecting those, drawn from the profile in connection 
with the true face, producing the miter line, as shown 
by E 2 . 

By inspection of the plan where a portion of the 
bed mold is shown it will be seen that the miter of the 
bed molding around the octagon at E is regular — that 
is, its miter line does not coincide with the line of the 
hip D E. If the profile of the bed molding in the 
vertical section, and also the profile of the bed mold- 
ing as shown in the plan, be divided into any equal 
umber of parts, points may be dropped from the 



Pattern Problems. 



231 




Fig. 415.— Plan, Elevation, True Face and Part of Pattern. 



Fig. 417.— True Face of Octagonal Side and Part of Pattern. 



Patterns for a Hip Molding Mitering Against the Bed Molding of a Deck Cornice on a Mansard Roof which is Square at the Base and 

Octagonal at the Top. 



232 



Tlic New Metal Worker Pattern Book. 



profile of the plan on to the miter line E, and thence 
carried downward and intersected with horizontal lines 
from the corresponding points of the bed molding in 
section, also shown at E 2 , thus giving the appearance of 
the miter between the two arms of the bed mold behind 
their intersection with the hip roll. The vertical 
lines from the miter E of the plan have not been car- 
ried, in the engraving, further than E 1 , where they are 
intersected with lines from corresponding points from 
the profile at B of the elevation, thus showing how 
the operation is performed. This has been done to 
avoid a confusion of lines at E\ Having obtained 
this line in the true face, the point where it crosses 
the miter line between the hip mold and bed mold 
previously obtained at E 2 must be noted. A line from 
this point of intersection must then be carried parallel 
to the line of the molding in the true face, back to the 
profile of the hip, and there marked, as shown by the 
figure 7£. The position of the point 7-J should now 
be marked upon the section of the hip molding previ- 
ously obtained at O 1 in Fig. 416. So much of the 
profile as exists between 1 and 7% in the true face is 
used in obtaining the stretchout of this part of the 
pattern. The remaining portion of the stay — namely, 
from 7£ to 14 — is afterward used for the true face of 
the octagonal side for the remainder of the pattern. 

At right angles to the line of the molding in the 
true face lay off a stretchout equal to that portion of 
the profile thus used, as shown by P 1ST, through the 
points in which draw measuring lines in the usual 
manner. Place the T-square at right angles to the 
lines of the molding in the true face, and, bringing it 
against the several points in the miter line between 
the hip and bed molding at E 2 , cut corresponding 
measuring lines drawn through the stretchout. Then 
a line traced through these points, as shown by S T, 
will be the miter line for that portion of the pattern 
corresponding to the part of the profile thus used. 

For the other half of the hip molding, being that 
portion which lies on the face of the transition piece, 
another operation must be gone through. Construct a 
section of the roof corresponding to the line F G in 
the plan. At any convenient point lay off F 1 C 1 in 
Fig. 415, equal in length to F G. From the point C 
erect a perpendicular, C B 3 , in length equal to C B of 
the section. Connect F' and B 3 . Then F 1 B 3 is the 
length of the transition side of the roof through that 



portion corresponding to F G of the plan. It will be 
well to add to this at B 3 a section of the bed mold as 
it appears in the section below, thus establishing the 
true relation between it and the transition side of the 
roof. By means of this section and the plan, construct 
a true face of one-half the transition side of the 
roof, by means of which to obtain the miter of the re- 
maining portion of the roll. To do this first redraw 
the section B 3 F', jnacing the line of the roof in a ver- 
tical position, as shown by B 4 F 2 , Fig. 417, from the 
points in which project horizontal lines, as shown to 
the right, upon each of which set off from an assumed 
vertical line the width of the roof as given in the plan. 
Thus make G' E 4 equal to G E of the plan, and F 3 
D 4 equal to F D. Connect D 4 and E 4 . Then G' E 4 D 4 
F 3 is the true face of that portion of the roof repre- 
sented by G E D F in the plan. 

In connection with the vertical section just de- 
scribed place so much of the stay as was not used for 
the pattern already delineated, and in the elevation of 
the transitional face of the roof place a corresponding 
portion of the profile, as shown, each of which divide 
into the same number of spaces. From the points 
thus obtained carry lines parallel to the lines of the 
respective views of the part, those in the vertical sec- 
tion cutting the bed molding, and those in the eleva- 
tion being produced indefinitely. From the points in 
the bed molding of the vertical section carry lines 
horizontally, intersecting those drawn from the profile 
in the elevation, thus establishing the miter line, as 
indicated at E 4 . At right angles to the line D 4 E 4 set 
off a stretchout of the profile, as shown by B P 2 , 
through the points in which draw the usual measuring 
lines. With the T-square placed parallel to this 
stretchout line, or, what is the same, at right angles to 
the line D 4 E 4 , and being brought successively against 
the points in the miter line at E 4 , cut corresponding 
measuring lines, as shown. Points also are to be car- 
ried across, in the same manner as described, corre- 
sponding to the bottom of the apron or fascia strip in 
connection with the bed molding. Then a line traced 
through these points, as indicated by the line drawn 
from U to T 1 , will be the pattern of the other half of 
the hip molding. By joining the two patterns thus 
obtained upon the dividing line of the stay, correspond- 
ing to P T of the first piece or P 2 T 1 of the second 
piece, the pattern will be contained in one piece. 



Pattern Problems. 



233 



B 3 (DUPLICATE OF PAGE 231.) 




Fig. 415.— Plan, Elevation, True Ftce and Part of Pattern. 



Fig. 417.— True Face of Octagonal Si le and Part of Pattern. 



Patterns for a Hip Molding Mitering Against the Bed Molding of a Deck Cornice on a Mansard Roof which is Square at the Base and 

Octagonal at the Top. 



234 



Tlie New Metal Worker Pattern Book. 




Fig. 418. — The Patterns for the Miter at the Bottom of a Hip Molding on a Mansard Roof Which is Octagon at the Top and Squart, 

at the Bottom. 



Pattern Problems. 



234a 



PROBLEM 109. 

The Patterns for the Miter at the Bottom of a Hip Molding: on a Mansard Roof which is Octagon at the 

Top and Square at the Bottom. 



Let LDB' in Fig. 418 be the plan of the roof at 
the base, and Pi C G 1 the plan at the top of the portion 
here made use of for the purpose of demonstration. 
The section A B P in Fig. 419 shows the pitch of the 
roof and of the wash at the bottom upon which the 
hip molds miter. The pattern then required will be 
the shape of the hip molding to miter against this wash. 
Since the roof is square at the base and octagonal at 
the top, the moldings covering the two converging 
hips, represented by R D and C D, of Fig. 418, will 
unite and become a single profile at D. But, since the 
two hip moldings join before the wash is reached, the 
pattern will be modified to the extent of fitting the 
inner edge of one against the corresponding edge of 
the other. This condition of things is shown in the 
elevation, Fig. 419, which is here introduced not for 
any use it may be in the operation of cutting the pat- 
terns, but for more clearly showing the nature of the 
problem. The elevation is drawn by means of inter- 
secting points from the section and the plan, viz. : D 5 
H' B 3 corresponds to I) H B' in the plan, and horizontal 
lines from the points A B in the section aie drawn, 
intersecting lines corresponding to the points already 
named. 

Let M L of the section represent one-half of the 
profile of the molding which is required to be fitted to 
the converging hips. The first step in the develop- 
ment of the patterns is in the construction of a section 
upon the line of one of the hips. Therefore, lay off 
D' C 2 equal to D C of the plan, and from C 2 erect a 
perpendicular, C 2 A 1 , in length equal to C A of the 
original section. Connect A 1 D 1 . Then A' C 2 D 1 is a 
section of the roof as it would appear if cut through 
on the line D C 1 of the plan, and A 1 D 1 is the pitch of 
the hip. In order to locate the profile of the hip mold- 
ing upon this section in correct position take any two 
points, as G and GF, on either side of the angle and 
equidistant from it. From Gr and G' carry lines paral- 
lel to C A 1 , producing them until they cut the hori- 
zontal line drawn through A 1 at the top of the section, 
as shown by the point K 1 and F'. From K 1 and F 1 
draw lines parallel to the pitch line A 1 D'. At any 



convenient place in A 1 D' establish the point 0, through 
which draw a line at right angles to A 1 D', of conven- 
ient length. From the intersection of the line just 
drawn through with the line from K 1 set off the 
distance K G in the plan, locating the point G 2 . From 
F 2 , on a continuation of the line F 1 F 2 , set off a dis- 
tance equal to F G 1 in the plan, as shown at G 3 . Con- 
nect the point with G 2 and G 3 . Then the point of 
the newly constructed profile will represent the corner 
of the hip, to which the fascias and roll may be added 
according to the requirements of the original designs. 

The nexL step will be the construction of a "true 
face" of one of the normal sides so as to obtain the 
correct angle of the hip with the base line. To do 
this, transfer the section A B P of Fig. 419 to a con- 
venient place below the plan, bringing the line A B to 
a vertical position, keeping the angle between the sur- 
face of the roof and the wash the same, as shown at 
A 3 B 2 P 2 . From the points in this section project lines 
horizontally to the left, intersecting them with lines 
from the plan corresponding with the points used in 
the section. Then B 3 D 3 P 3 Z will represent a true 
face of the roof with an elevation of the wash to cor- 
respond. Place a portion of the stay in this true 
face, locating it so that the point O 1 , which corre- 
sponds to of the hip section, shall fall upon the angle 
of the roof. Divide it into any convenient number of 
spaces, numbering them in the usual manner. , From 
these points drop lines parallel to B 3 D 3 indefinitely 
through the face of the wash. Place also a part of the 
profile of the hip molding (greater than one-half) in 
proper position in the vertical section. From the 
points in this profile drop lines cutting the wash P 2 B 2 . 
From the points thus obtained carry lines horizontally, 
Crossing the true face, intersecting them with lines of 
corresponding numbers previously drawn. A line 
traced through the intersection of these points will 
give the elevation of the miter in the true face, all as 
shown by S T U. 

Note the point where this miter line crosses the 
miter line of the wash P 3 D 3 , which intersection carry 
back upon the profile L 2 M 2 , which in this case will 



2?Ab 



New Metal Worker Pattern Book. 
(DUPLICATE OF PAGE 234.) 




Fig. 4I8. — The Patterns for the Miter at the Bottom of a Hip Molding on a Mansard Roof Which is Octagon at the Top and Square 

at the Bottom. 



Pattern Problems. 



235 



correspond to the point 8. Locate the point S on the 
first section of the hip obtained near 0, as shown, and 
use the remainder of profile 8 to 14 for another opera- 
tion. Lay off a stretchout of the entire profile of the 
hip molding, as shown by W V, through the points in 
which draw the usual measuring Hues. With the 




Fig. 419.— Section and Elevation of the Miter at the Bottom of a Hip Molding 
on a Mansard Roof Which is Octagon at the Top and Square at the Bottom. 




True Face of Octagon 



Fig. 420.— Miter between the Inner Edges of the Hip Moldings at the Bottom. 



hip mold it will be necessary first to construct a true 
face of the octagon side of the roof. To do this, 
obtain a diagonal section of the roof corresponding to 
the line D E in the plan, viz. : Lay off D 2 E 1 equal to 
D E of the plan, and from E 1 erect a perpendicular, 
E 1 A 2 , equal to C A of the section in Fig. 419. Con- 
nect A 2 and D\ Then A 2 D 2 is the 
length of the diagonal face of the roof 
measured on the line D E of the plan. 
Upon any convenient straight line lay off 
D 4 A 4 in Fig. 420, in length equal to D 1 
A 2 , and from A 4 set off, at right angles to 
it, A' C\ in length equal to EC of the 
plan. Then D 4 A* C 3 shows in the flat one- 
half of the diagonal face of the roof, or 
what is represented by D E C 1 in the plan. 
At right angles to D 4 C 3 draw the remain- 
ing portion of the stay not used in con- 
nection with the true face, placing it in 
such a manner that the point O 3 , corre- 
sponding to of the hip section, shall fall 
upon the line D 4 C 3 , which represents the 
angle of the hip. Through the point 8 of 
the section L 5 M 7 , corresponding to 8 of the 
section L 2 M 2 of Fig. 418, draw a line 
parallel to D' C 3 , as shown by S 2 Y'. Then 
S 2 Y 1 corresponds to S Y of the true face 
in Fig. -118. 

Space the profile L B M' into the same 
parts as used in laying off the stretchout 
W V, and through the points draw lines 
parallel to D 4 C 3 , cutting the line S 2 A', 
which, being the center line of the octag- 
onal side of the roof, is also the miter 
line between the two arms of the hip mold- 
ing. From the points of intersection 
in the line D 4 A 4 , at right 



T-square placed at right angles to the lines of the hip, 
as shown in the true face, and brought against the 
points in the miter line S T IT, cut so many of the 
measuring lines drawn through the stretchout W V as 
correspond to those points. By this means that por- 
tion of the pattern shown by S 1 T 1 U' will be obtained. 
For the portion of the pattern corresponding to 
the part of the profile which miters against the other 



jles to S 2 

Y', draw lines cutting S 2 Y', giving the 
points marked 8, 9, 10, 11, 12, 13 and 14. For con- 
venience in using one stretchout for the entire pat- 
tern, transfer these points to the line S Y of the true 
face in Fig. 418, from which, at right angles to S Y, 
draw lines cutting the corresponding measuring lines 
of the stretchout. Then a line traced through these 
points of intersection, as shown from S 1 to X, will com- 
plete the pattern. 



236 



Tlie New Metal Worker Pattern Book. 



PROBLEM no. 



Patterns for the Fascias of a Hip Molding: Finishing: a Curved Mansard Roof which is Square at tht 

Base and Octag-onal at the Top. 



The conditions involved in this problem do not 
differ greatly from those given in Problems 80, 81 and 
82, near which it should properly be classed. In this 
case, however, the profile of an entire roof is under 
consideration instead of that of a simple molding or 
vase, but the problem is here introduced as being 
closely related in feature to several of the foregoing- 
problems. 

C D E F of Fig. 421 represents the plan of the 
roof at its base, while V G II W represents the plan at 
the top. It will be seen that the roof is nearly square 
at the foot of the rafters and octagonal at the top. The 
same conditions may arise where the corners of the 
roof are chamfered, starting at nothing at the bottom 
and increasing to a considerable space at the top, with- 
out reference to forming an octagon. D G H E in the 
plan represents a chamfer or transition piece in the con- 
struction of a roof which, as above described, is square 
at the base and octagonal at the top. This part is rep- 
resented m elevation by D 2 G" FT E\ The elevation is 
introduced here not for any use in pattern cutting, but 
simply to show the relation of parts. In the sectional 
view of the roof A B the outer line B represents 
the surface of the fascias of which the patterns are re- 
quired, the inner curve showing the line of the roof 
boards and the depth of the sink strips. As it is in the 
plan that the miter lines are shown it will be necessary 
to develop the pattern from the plan. Assuming then 
that one of the square sides, as E F W H, is to be 
done first, it will be necessary to place a profile so that 
its projection A shall lie across this part of the roof, 
all as shown by O 1 A 1 B'. 

Divide the profile 0' B 1 into any convenient num- 
ber of equal spaces, and from the points of division 
drorj lines parallel to E F, the side of the roof, cutting 
the miter line E H. Upon any line at right angles to 
this side of the roof, as O 2 B', lay off a stretchout 
through the points, in which draw the usual measuring 
lines. Cut these measuring lines by lines drawn verti- 
cally from the points in E H. Then a line traced 
through these points of intersection, as shown by E 3 
FT, will be the line of the pattern corresponding to the 



line E II in the plan. The width of the flange or fascia 
forming the hip finish may be obtained as described in 
Problem 6, and the corner piece drawn in to agree with 
the original design, as shown by S' T 1 

If it is desirable to produce an elevation of this 
angle of the roof it can be done by dividing the profile 
B by the same points as were used in dividing O 1 B 1 , 
from which horizontal lines can be drawn to the left 
intersecting with the lines of corresponding number 
previously erected from the miter line EH. A line, 
E a IP, drawn through the points of intersection will 
with D 2 Gr s give the correct elevation of the transition 
side. 

For the pattern of this side it will be necessary to 
first construct a section upon its center line, P R of the 
plan. At any convenient place outside of the plan 
draw a duplicate of P R parallel to it, as shown by 
P 1 A 1 , and from the point A 1 erect a perpendicular, 
A 1 B', in length equal to A B of the original section. 
In A' B 1 set off points corresponding to the points in 
A B, and through them draw horizontal lines, as shown. 
Place the T-square parallel to A 1 B 1 , and, bringing it 
against the points in E II previously obtained from the 
profile O l B 1 , cut corresponding measuring lines. Then 
a line traced through these points of intersection, as 
shown by B' P 1 , will complete the diagonal section 
corresponding to P R in the plan. From this diagonal 
section take a stretchout, which lay off on the straight 
line corresponding to P R produced, all as shown by 
P 2 B 3 . Through the points in P 2 B 3 draw the usual 
measuring lines. With the T-square placed parallel to 
this stretchout line, and brought successively against 
the points in E II, cut the measuring lines, as shown. 
Then a line traced through these points of intersection, 
as shown by E 1 to FT, will be one side of the required 
pattern. In like manner, having transferred points 
from E H across to the corresponding line D G, cut the 
measuring lines from it, which will give the other side 
of the required pattern. The width of fascias (whose 
intersection forms a panel in this case) may be obtained 
as suggested above and as given in Problem 6. 

In locating the points IT 1 and M 1 of this pattern it, 



Pattern Problems, 



237 




Fig. 4S1-— Patterns for the Fascias of a Hip Molding Finishing a Curved Mansard Roof Which is Square at the Eaves and 

Octagonal at the Top. 



238 



TJie New Metal Worker Pattern Book. 



is desirable for the sake of design that they be, when 
finished and in position, at the same vertical distance 
below the cornice as are the points S and T on the 
square sides of the roof. To accomplish this it will be 
necessary to go back to the points S 1 and T', in the first 
pattern obtained, and from them carry lines back into 
the dtretchout line 0" B 1 , where they are numbered 
10-J- and 11^-. Their positions may now be transferred 
by means of the dividers to the normal profile B, 
where their vertical hights can be measured on the line 
A B, as shown, and transferred again to the vertical 
line A 1 B 1 of the diagonal section. It is only neces- 
sary now to carry them across, as shown, to the profile 
P 1 B', where their distances from adjacent points may be 
measured by the dividers and placed upon the stretch- 
out line P 2 B 3 . By similar means the appearance of 
this panel bof h in the plan and in the elevation may be 



completed if so desired, all of which will be made clear 
by inspection of the drawing. 

In the case of very large roofs, where the develop- 
ment of a profile or a pattern to the full size would be 
impracticable, it is possible to perform the work to a 
scale of 1-J- or 3 inches to the foot; after which full 
size patterns of parts of convenient size may be ob- 
tained by multiplying their various dimensions by 8 
or 4. 

As the patterns for the roll, usually finishing 
the hip, are properly included under the head of 
Flaring Work, which subject is treated in the fol- 
lowing section of this chapter, they will not be 
given here. The radii from which they can be ob- 
tained, however, may be derived from the diagonal 
section in the manner described in the following 
problem. 



PROBLEM in. 



To Obtain the Curves for a Molding; Covering- the Hip of a Curved Mansard Roof. 



The method of obtaining the pattern of the fascias 
of a molding covering a curved hip has been given in 
Problem 6. As it is necessary in obtaining the pat- 
terns of the molded portion or roll, that the curve of the 
hip should be established, this problem really consists 
of developing from the normal profile of the roof a 
profile through the hip, or, in other words, a diagonal 
section of the mansard. 

Let A E B in Fig. 422 represent the plan of a 
mansard roof or tower, the elevation of which is shown 
by H K, over the hip of which a molding of any given 
profile is to be fitted, in this case a three-quarter bead, 
the diagonal line E F in the plan representing the 
angle, of the hip as it would appear if viewed from the 
top. At any convenient point parallel to E F, and 
equal to it, draw E 1 F 1 , and from F 1 erect a perpendicu- 
lar, F 1 K 1 , in length equal to the vertical line in eleva- 
tion Gr K. Divide Gr K and F 1 K 1 into the same num- 
ber of equal spaces. From the points in G K draw 
lines cutting the profile H K, as shown, and from the 
points thus obtained in H K drop lines vertically, pro- 
ducing them until they cut the diagonal line E F of the 
plan, as shown. Through the points in F 1 K 1 draw 



measuring lines in the usual manner, and intersect them 
by lines erected perpendicularly to E F from the points 
therein. Then a line traced through these points of 
intersection, as shown by E' K 1 , will be the profile to 
which the molding covering the hip is to be raised. 

Inasmuch as in the usual process of mold raising 
all curves must be considered as segments of circles, 
to accommodate both the adjustment of the machine 
used and the describing of the patterns, the curved 
line E 1 K 1 just obtained must be so divided that each 
section or segment will approach as nearly as possible 
an arc of a circle. In this case the section from E 1 to 
L will be found to correspond to an arc struck from 
a center, M, while the section from L to K' corresponds 
to an arc struck from a center not shown in the engrav- 
ing, but which will be found by the intersection of the 
lines L N" and K 1 N 1 produced. 

In the lower part of Fig. 423 is shown an enlarged 
section of the hip molding, including the fascias, as it 
would appear at the bottom of the hip, and above it 
another section taken at the top, which has been de- 
rived from the normal section or section at the bottom 
by the method used and explained in Problems 105, 



Pattern Problems. 



239 



106 and 107, previously demonstrated. A dotted 
reproduction of the lines of the upper section is 



the roll require trimming after being raised so that the 
roll may have an equal projection throughout its course. 



SECTION AT TOP 



Fig. 423.— Enlarged Sections Through Hip Finish at 
Top and Bottom, Showing Change in Flare of Fascias. 




A F ft ' V 

Fig. 422.— Diagonal Section of a Curved Mansard Roof Obtained for the Purpose of Mold Raising. 



placed here to show the change in the flare that takes 
place between fascias in going from the bottom to the 
top of the hip, thus showing that the outer edges of 



Methods of obtaining the patterns of curved mold- 
ings will be found in the following section of this 
chapter. 



240 



Tlie New Metal Worker Pattern Book. 

SECTION 2. 

Regular Tapering; Forums 

(FLARING WORK.) 



It will be well to place before the reader here a 
clear statement of the class of problems he may ex- 
pect to meet with under this head. It will include 
only the envelopes of such solid figures as have for a 
base the circle, or any figure of equal or unequal sides 
which may be inscribed within a circle, and which 
terminate in an apex located directly over the center 
of the base. 

According to the definition of an inscribed poly- 
gon (Def. 66), its angles must all lie in the circum- 
ference of the same circle. So the angles or hips of a 
pyramid whose base can be inscribed in a circle must 
lie in the surface of a cone whose base circumscribes 
its base and whose altitude is equal to that of the 
pyramid. Therefore the circle which describes the 
pattern of the base of the envelope of such a cone will 
also circumscribe the pattern of the base of the pyra- 
mid contained within it. The envelopes of such solids, 
therefore, as scalene cones, scalene pyramids and 
pyramids whose bases cannot be inscribed within a 
circle are not adapted to treatment by the methods 
employed in this section. Even the envelope of an 
elliptical cone cannot be included with this class of 
problems because it possesses no circular section upon 
which its circumference at any fixed distance from 
the apex can be measured. 



In this connection it is proper to call attention to 
the difference between a scalene cone and a right cone 
whose base is oblique to its axis. According to Defi- 
nition 96, a scalene cone is one whose axis is inclined 
to the plane of its base, and according to Definition 94 
the base of a cone is a circle. As any section of a 
cone taken parallel to its base is the same shape as its 
base, any section of a scalene cone taken parallel to 
its base must be a circle, and any section taken at 
right angles to its axis could not, therefore, be a circle, 
but would be elliptical. Again, as any section of a 
right cone (Def. 95) at right angles to its axis is a 
circle, if its base be cut off obliquely, such base would, 
according to Definition 113, be an ellipse. There- 
fore, since its horizontal section is a circle, its 
envelope may be obtained by methods employed in 
this section. (See Problem 136.) And since the sec- 
tion of a scalene cone taken at right angles to its axis 
is an ellipse, the scalene cone becomes virtually an 
elliptical cone with an oblique base — that is, with a base 
cut off at such an angle as to produce a circle — and, as 
stated above, cannot be included in this section. 

The principles governing the problems of this 
section are given in Chapter V, beginning on page 79, 
which the reader will find a great help in explaining 
anything which he may fail to understand. 



PROBLEM H2. 

The Envelope of a Triangular Pyramid. 



Let A B C of Fig. 424 be the elevation of the 
pyramid, and E F Gr of Fig. 425 the plan. From the 



the point K erect K H, perpendicular to F K and equal 
in length to the hight of the pyramid, as shown by 






Fir. 424.— Elevation. 



Fig. 435.— Plan. 
The Envelope of a Triangular Pyramid. 



Fig. 436.— Pattern. 



center K draw the lines E K, F K and Gr K in the plan, 
representing the angles or hips of the pyramid. From 



A D of the elevation. Draw the hypothenuse F H, 
which then represents the length of the corner lines. 



Pattern Problems. 



241 



From any point, as L of Fig. 426, for center, with 
radius equal to F H, describe the arc M N I indef- 
initely, and draw L M. From M set off the chord M 
N, in length equal to the side F G of the plan. In like 



manner set off N and I respectively, equal to G E 
and E F of the plan. Connect I and L, as shown, 
and draw L and L ST. Then L I N M is the 
pattern sought. 



PROBLEM 113. 

The Envelope of a Square Pyramid. 



Let E A C of Fig. 427 be the elevation of the 
pyramid, and F H K L of Fig. 428 the plan. The 
diagonal lines F K and L II represent the plan of the 
angles or hips, and G a point corresponding to the 
apex A of the elevation. From the apex A drop the 



or circumscribe the pattern, as shown in the diagram 
From any center, as M, Fig. 429, with a radius equal 
to A D, describe an arc, as P E O S N, indefinitely 
and draw M P. From P, on the arc drawn, set off 
a chord, P R, in length equal to one of the sides of 




Fig. 437.— Elevation. 




Fig. 428.— Plan. 
The Envelope of a Square Pyramid. 




line A B perpendicular to the base E C. Prolong E 
C in the direction of D, making B D equal to G F, one 
of the angles of the plan. Connect D and A. Then 
A D will be the slant hight of the article on one of the 



corners, and the radius of an 



arc 



which will contain 



the pyramid shown in the plan. From R set off 
another chord, R 0, in like manner, and repeat the 
same operation, obtaining S and S X. Draw the lines 
M N, M S, M and M R. Then MNSOEP will 
be the required pattern. 



PROBLEM 114. 

The Envelope of a Hexagonal Pyramid. 



Let H G I of Fig. 430 represent the elevation of 
a hexagonal pyramid, of which DFCL B E of Fig. 
431 is the plan. The first step is to construct a section 
on a line drawn from the center of the figure through 
one of its angles in the plan, as A B. From the center 
A erect A X perpendicular to A B, making it equal to 



the straight hight of the article, as shown in the eleva- 
tion by G K. Draw the hypothenuse B X. Then X 
represents the apex and XB the side of a right cone, the 
plan of the base of which, if drawn, would circum- 
scribe the plan of the hexagonal pyramid. From any 
convenient center, as X' of Fig. 432, with X B of 



242 



The New Metal Worker Pattern Book. 



Fig. 431 as radius, describe an arc indefinitely, as 
shown by the dotted line. Through one extremity of 
the arc to the center draw a line, as shown by D' X ! . 



B 1 L 1 in the arc thus obtained draw lines to the center, 
as shown by E' X', B' X 1 , etc., which will represent 
the angles of the completed shape, and serve to locate 




l\ 












\ 
1 
1 
1 


w 












1 


\\ 












/ 


\\ 












/ 
II 


E 1 


\\ 

B' 


^^; 








V 



Fig. 430— Elevation. 



Fig. 431.— Plan. 



Fig. 432.— Pattern. 



The Envelope of a Hexagonal Pyramid. 



With the dividers set to a space equal to any side of the 
plan, as D E, commencing at D 1 , setoff this distance on 



the bends to be made in process of forming up. 
Then X 1 D 1 E' B 1 L' C* F 1 D 2 will be the complete 



the arc six times, as shown. From the several points E 1 | pattern. 

PROBLEM 115. 

The Envelope of the Frustum of a Square Pyramid. 



In Fig. 433, let G H K I be the elevation of the 
article, C A E D the plan of the larger end and L M 



Construct a diagonal section on the line A P as fol- 
lows : Erect the perpendicular P F, making it equal 



\ 


M 




/ 


V 




1 


K. 












/ 



\ 
\ 

K. \ 



\ 



Fig. 433— Plan and Elevation. 




Fig. 434.— Pattern. 
The Envelope of the Frustum of a Square Pyramid. 



1ST the plan of the smaller end. Produce the hip 
lines C L, A M, etc., in the plan to the center P. 



to the straight hight of the article, as shown by E K 
of the elevation. Likewise erect the perpendicular 



Pattern Problems. 



243 



M B of the same length. Draw F B and A B. Then 
P A B F is the diagonal section of the article upon 
the line P A. Produce A B indefinitely in the direc- 
tion of X, and also produce P F until it meets A B 
extended in the point X. Then X is the apex of a 
right cone and X A the side of the same, the base of 
which, if drawn, would circumscribe the plan C A E D. 
Therefore, from any convenient center, as X 1 of Fig. 
434, with X A as radius, describe the arc C D' E' A 1 
C 2 , and from the same center, with radius X B, draw 



the arc L' X' : M' L 2 , both indefinitely. Draw C 
X 1 , cutting the smaller arc in the point L 1 . Make the 
chord C D 1 equal in length to one side, U D, of the 
plan, and D 1 E 1 to another side, D E, of the plan, 
and so on, until the four sides of the base have been 
setoff. Draw D' X', E' X', etc., cutting the arc 
L 1 L 2 'in the points N 1 , 1 , etc. Then D 1 N 1 , E 1 O 1 
and A' M 1 will represent the lines of the bends in 
forming up the pattern. Draw the chords L' N 1 , X : O 1 , 
etc., thus completing the pattern. 



PROBLEM 116. 



The Envelope of the Frustum of an Octagonal Pyramid. 



H N 




O' p. 5 

Fig. 437.— Pattern. 

The Envelope of the Frustum of an Octagonal Pyramid. 



Fig. 435 shows the elevation and Fig. 43G the 
plan of the frustum of an octagonal pyramid. The 
first step in developing the pattern is to construct a 
diagonal section, the base of which shall correspond to 
one of the lines drawn from the center of the plan 
through one of the angles of the figure, as shown by 
Gr B. Erect the perpendicular G- C equal to the 
straight hight of the frustum, as shown by X M of the 
elevation, and at b erect a perpendicular, b A, of like 
length. Draw B A and A C. Then G B A C is a 
section of the article as it would appear if cut on the 
line Gr B. Produce B A indefinitely in the direction 
of X, and likewise prolong Gr C until it intersects B A 
produced in X. Then X is the apex and X B the 
side of a right cone, the plan of which, if drawn, 
would circumscribe the base of the frustum. From 
any convenient center, as X', Fig. 437, with radius 
X B, describe an arc indefinitely, as shown by the 
dotted line E' E 2 of the pattern, and from the same 
center, with X A for radius, describe the arc e' e 2 of 
the pattern. Through one extremity of the arc E 1 Ej 
to the center draw a straight line, as shown by E 1 X 1 
cutting the smaller arc in the point e'. Set off on the 
arc E 1 E 2 spaces equal to the sides of the plan of the 
base of the article and connect the points by chords. 
Thus make E 1 P 1 of the pattern equal to E P of the 
plan, and so on. Also from these points in the arc 
draw lines to the center, cutting the arc e 1 e 2 , as shown. 
Connect the points thus obtained in this arc by chords, 
as shown by e 1 p\ p l d\ d 1 o\ etc. Then e' E' E 2 e 2 will 
be the pattern sought. 



244 



Tlte New Metal Worker Pattern Book. 



PROBLEM 117. 

The Envelope of the Frustum of an Octagonal Pyramid Having: Alternate Long- and Short Sides. 



In Fig. 438, let IMBNOPKLbe the plan of 
the article of which G H F E is the elevation. The 
first thing to do in describing the pattern is to construct 
a section corresponding to a line drawn from the center 



Produce S R and B A until they meet in the point X. 
Then X is the apex and X B is the side of a cone, the 
base of which, if drawn, would circumscribe the plan 
• of the article. From any convenient center, as X 1 , 
Fig. 439, with radius equal to X B, describe an arc. 
as shown by M 1 M 2 . Draw X 1 M 1 as one side of the 
pattern. Then, starting from M 1 , set off chords to the 
arc, as shown by M 1 B 1 , B 1 X 1 , etc., equal to and corre- 
sponding with the several sides of the article, as shown 
by M B, B N, etc., in the plan. From these points, 
B 1 , N 1 , etc., in the arc, draw lines to the center X 1 . 




o' p' 

D f 

Fig. 438.— Plan and Elevation. Fig. 439— Pattern. 

The Envelope of the Frustum of an Octagonal Pyramid Having Alternate Long and Slwrt Sides. 



to one of the angles in the plan, as S B. At S erect 
the perpendicular S R, in length equal to the straight 
hight of the article, as shown by C D of the elevation. 
Upon the point b erect a corresponding perpendicular, 
as shown by b A. Draw R A and A B. Then B A 
R S is a section of the article taken upon the line S B. 



From X 1 , with X A as radius, describe an arc cutting 
these lines, as shown by to 1 m\ Connect the points of 
intersection by straight lines, as shown by m l b 1 , b 1 n', n l o 1 , 
etc. Then m l m 2 M 2 M 1 will be the pattern sought, 
and the lines B 1 V, X 1 n\ etc., will represent the lines 
of bends to be made in forming up the article. 



PROBLEM 118. 

The Pattern of a Square Spire Miteringf Upon Four Gables. 



In Fig. 440, let B F H C be the elevation of a 
square spire which is required to miter over four equal 
gables in a pinnacle, the plan of which is also square. 



Produce F B and H C until they meet in A, which 
will be the apex of the pyramid of which the spire is a 
section. Draw the axis A G, and at right angles to it, 



Pattern Problems. 



245 



from the lowest point of contact between the spire and 
the gable, as F, draw F G. Then F G will represent 
the half width of one of the sides of the pyramid at the 
base, and A F will represent the length of a side 
through its center. From any convenient point, as A 1 



spaces of the extent of Gr' Gr 2 , as shown by G 2 g, g g' and g 2 
0. Draw g 1 A 1 , g A 1 and A 1 . Make A 1 B' equal to A 
B of the elevation, and through B 1 draw a perpendicu- 
lar to A' F 1 , as shown. Draw lines corresponding to it 
through the other sections of the pattern. Make A 1 D 1 




Fig. 440.— Elevation of Spire. 




Fig. 441.— Pattern. 
The Pattern of a Square Spire Mitering Upon Four Gables. 



in Fig. 441, draw A 1 F 1 , in length equal to A F. From 
F 1 set off, perpendicular to A' F 1 , on each side a space 
equal to F G of the elevation, as shown by F 1 Gr' and 
F' Gr 2 . From Gr 1 and G a draw lines to A 1 , as shown. 
From A 1 as center, and with A' G 2 as radius, describe 
an arc, as shown by G 2 0, in length equal to three 



equal to A D, and draw D 1 G 1 and D 1 G 2 . Set the 
compasses to G 2 D', and from G 2 and g as centers 
describe arcs intersecting at d. Draw d g and d G 2 , as 
shown. Repeat the same operation in the other 
sections of the pattern, thus completing the required 
shape. 



PROBLEM 119. 

The Pattern of an Octagon Spire Mitering Upon Eight Gables. 



Let A G L in Fig. 442 be the elevation of the 
spire, and MOP the half plan. From the point G, 
which represents the lowest point of the angle or valley 
between the gables, to H, which represents the meet- 



ing of the valleys and ridges at T in the plan, draw 
the line G H, cutting the side A C extended in the 
point D. Draw any line, as A 1 D 1 in Fig. 443, upon 
which to construct the pattern. Make A 1 C 1 equal to 



240 



Tlie New Metal Worker Pattern Book. 



A C of the elevation, and A 1 D 1 equal to A D of the 
elevation. Through D 1 draw the horizontal line 1 0, as 




Draw A 1 and A 1 1. Set the dividers to A 1 1 as 
radius, and from A 1 as center describe the arc 1 8 in- 
definitely. Set the dividers to 1 0, and step off as 
many spaces on the arc as there are sides in the spire. 
Draw the lines A 1 2, A' 3, etc., to A 1 8, which represent 
the angles of the spire and the bends in the pattern. 
Draw C and C 1 1 in the first section of the pattern. 
Set the dividers to C 1 1, and from 1 and 2 as' centers 
describe intersecting arcs, as shown by C. In like 
manner describe similar intersecting arcs at the points 




Fig. 443.— Plan and Elevation of Spire. 



Fig. 413— Pattern. 



The Pattern of an Octagon Spire Mitering Upon Eight, Gables. 



shown. From D 1 set off D' 0, equal to E F of the ele- 
vation, and likewise set off D 1 1, of the same length. 



C 3 , C 4 , etc. Draw lines from these points to the points 
1, 2, 3,4, etc., as shown, thus completing the pattern. 



PROBLEM 120. 

The Pattern of an Octagon Spire Mitering Upon Four Gables. 



In Fig. 444, let B E Z U be the elevation of an octa- 
gon spire, mitering down upon four gables occurring 
upon a square pinnacle. Continue the side lines until 
they intersect in the apex A. Draw the center line A 
H, and from the point Gdraw G H perpendicular to the 
center line, showing half the width of one of the 
sides at the point Gr. By inspection of the elevation it 



will be seen that one-half the sides will be notched at 
the bottom to fit over the gables, while the others will 
be pointed to reach down into the angles or valleys be- 
tween the gables. 

To ascertain the correct length upon the center 
line of one of the pointed sides it will be necessary to 
construct a section through one of the valleys, for in- 



Pattern Problems. 



247 



stance, upon the line M' N' of the plan. Through the 
point J of the elevation draw the line J M at right 
angles to the center line, extending it to the left in- 
definitely, and from the point M set off upon this line 
the distance M N, equal to M 1 N 1 of the plan. Draw 




equal to A E of the elevation, etc. Through E 1 draw 
a perpendicular equal in length to the width of a side 
at the point E, or to twice Gr H, as shown in the ele- 
vation, placing one-half on each side from E', all as 
shown by L K. From L and K draw lines to A 1 . Erom 
A' as center, with A 1 L as radius, describe an arc, as 
shown by L U, indefinite in length. Set the dividers 
to the space L K, and step off spaces from L, as L Y, 
Y X, etc., until as many sides are set off as are required 
in the spire — in this case eight. Draw the lines A' Y, 
A 1 X, etc. From the point D', which, as will be seen 
by D in the elevation, corresponds to the top of the 
gable, draw lines to the points L and K, which gives 
the pattern for the notch in the first section. Set 
the dividers to L D' as radius, and from X and Y as 
centers describe arcs intersecting at W. Draw W 
X and W Y, and repeat this upon all the alternate 
sides throughout the pattern, as shown, locating the 




Fig. 444— Plan and Elevation of Spire. Tig. 445.— Pattern. 

The Pattern of an Octagon Spire Mitering Upon Four Gables. 



N P, and extend the side A E until it intersects this line 
at F. Then A F will be the correct length through 
the center line of one of the long sides. 

To describe the pattern first draw A 1 F 1 in Fig. 
445, equal to' A F of the elevation, and set off points 
on it corresponding to points in A F. Thus make 
A' B 1 equal to A B, A 1 D 1 equal to A D, and A 1 E 1 



points and P. For the pattern of the point, take 
a space between the points of the dividers equal 
to L F 1 , and from L and Y as centers describe small 
arcs intersecting at M, and draw M L and M Y. With 
the same radius repeat the operation upon the inter- 
mediate sides, establishing the points V, H and I, thus 
completing the pattern. 



248 



Tlie New Metal Worker Pattern Book. 

PROBLEM 121. 



Pattern for an Octagon Spire Mitering upon a Roof at the Junction of the Ridge and Hips. 



In Fig. 446, let A B C represent the front eleva- i vation let W N M K J II represent the octagon 
tion of the roof and A' a c C the corresponding plan. | spire and II' J' K' M' N' 0' the corresponding plan. 




PLAN. c ' c " PLAN 

Fig-. 446. — Plana and Elevations of Spire. 



C° Y° 

Fig. 447.— Section of Spire on Line of Hip. 



Pattern for an Octagon Spire Mitering Upon a Roof at the Junction of the Ridge and Hips. 



Also let D E F Gr be the side elevation of roof, and In the side elevation the spire is represented by P X 
A" a c' C" the corresponding plan. In the front ele- UTBQ, and in plan by P' Q' R' T' U' V. Only 



Pattern Problems. 



249 



the points in plans are designated by letters which 
represent similar points in the elevation. In order to 
draw the plans and elevations, including the miter 
lines, it may be found convenient to first construct the 
entire octagons, as indicated in the plans, and from 
these to project the elevations above, as shown. From 
the point u in front elevation, which represents the 
intersection of one of the rear angles of the spire with 
the roof, carry a line parallel with B F, cutting X p. 
From the point U draw the miter line U T, and from 
the points V U drop jaerpendiculars to plan, cutting 
X' V and X' U', from which points can be drawn 
the miter lines V U' T' of the plan. 

To obtain the miter line P' Q' R' of plan, from 
which is obtained the miter line Q R of side elevation, 
a diagram has been constructed in Fig. 447 which 
shows a section of spire and roof on the line C" X' of 
plan. To construct the diagram proceed as follows : 
Draw any line, as X° Y°. From X° set off the dis- 
tance X E of side elevation or W B of front elevation. 
The point E° represents the junction of hip and 
ridge. From X° set off the distance X S, and erect 
the perpendicular S° L°, making it in length equal to 
S P, and connect L° X°. Then X° S° L° is a duplicate 
of X S P. From X° set off the distance X Y and 
erect the perpendicular Y° C°, in length equal to 
X' C" of plan, and connect C° E°. Then L° X° repre- 
sents one side of spire, and C° E° the hip of the roof, 
and the point Q° the point of junction between the two. 

As the spire is a perfect octagon, the profile of 

- the side just constructed is in nowise different from 

either of those shown in the elevations. It simply has 

in addition the profile of one of the hips by means of 

which the correct hi°'ht of its intersection with the 



same (the point Q°) is determined. Draw Q° Z° paral- 
lel with C° Y°, and from the point X' of plan set off 
the distance Z° Q° of diagram, as shown by X' Q'. 
Connect R' Q' and Q' P'. From the point Q' in plan 
carry a line parallel with the center line X X', cutting 
the hip line D E at Q. Draw Q R, which shows the 
miter line in side elevation. From the point Q can be 
drawn the line Q J, cutting the hip lines A B and B C 
in front elevation at the points J and X, and the miter 
lines J K and M N drawn. The points K M in front 
elevation correspond with the points K' M' of plan. 

For the pattern proceed as follows : Draw I x of 
Fig. 448, equal to P X of side elevation, and from I 
erect the perpendiculars I p and I p', equal in length 
to L' P' of plan or L M of front elevation. From p 
and p draw lines to x, as shown. From x as center, 
with x p as radius, describe an arc, as shown by p' e, 
indefinite in length. Set the dividers to the distance 
p p and step off spaces from p, as p r, r t, etc., un- 
til as many sides are set off- -as are desired to be shown 
in one part of the pattern. For convenience in de- 
scribing the pattern draw the lines x r, x t, x c. Con- 
nect c and e and make c d equal to p I and draw x d. 
Bisect p r and draw x a, and from x, onxa, set off the 
distance X° Q° of Fig. 447, locating the point q. Draw 
p q and q r. From x, on x d, set off the distances X 
V and X U of side elevation, locating the points v and 
i. Through i draw a perpendicular cutting x c and 
x e in the points u and u', then draw u' v v u and 
u t. Then x p p q r t u v u' is the pattern for part of 
spire shown on plan by X' I P' Q' R' T' U' V P. 

Fig. 448 shows a little more than half the full 
pattern, which will be readily understood by a com- 
parison of reference letters. 



PROBLEM 122. 



The Envelope of a Right Cone. 



In Fig. 449 let A B C be the elevation of the 
cone and D E F the plan of the same. To obtain 
the envelope set the compasses to the space B A, or 
the slant hight of the cone, as a radius, and from 



any convenient point as center, as B 1 of Fig. 450, 
strike an arc indefinitely. Connect one end of the 
arc with the center, as shown by A 1 B 1 . 

With the dividers, using as small a space as i; 



250 



Tlie New Metal Worker Pattern Booh. 



convenient, step off the circumference of the plan 
D E F, counting the spaces until the whole, or exactly 




Fig. 449.— Plan and Blevation. 



B 1 A 1 C will be the pattern for the envelope of the 
cone ABC. 

It is not necessary that all of the spaces used in 
measuring the circumference of the plan should be 
equal. It frequently happens that when the space 
assumed between the points of the dividers has been 
stepped off upon the circumference of the base, a space 
will remain at the finish smaller than that originally 




Fig. 450.— Pattern. 



The Envelope of a Right Cone. 



one half, is completed, as shown in the upper half of 
the plan. Then set off on the arc A 1 C l of the pattern, 
commencing at A 1 , the same number of spaces as is 
contained in the entire circumference of the plan. 
Connect the last point C 1 with the center B 1 . Then 



assumed. In that case the required number of full 
spaces can be stepped off upon the arc of the pattern, 
after which the remaining small space may be added, 
thus completing the correct measurement of the 
pattern. 



PROBLEM 123. 



The Envelope of a Frustum of a Right Cone. 



The principle involved in cutting the pattern for 
the frustum of a cone is precisely the same as that for 
cutting the envelope of the cone itself. The frustum 
of a right cone is a shape which enters so extensively 
into articles of tinware that an ordinary flaring pan, an 
elevation and plan of which are shown in Fig. 451, has 
been engraved for the purpose of illustration. An in- 



spection of the engraving will show that C D, the top 
of the pan, is the base of an inverted cone, its apex B 
being at the intersection of the lines D and C A 
forming the sides of the pan ; and that A D is the top 
of the frustum or the base of another cone, A B, 
which remains after cutting the frustum from the orig- 
inal cone. For the pattern then proceed as follows : 



Pattern Problems. 



251 



Through the elevation draw a center line, K B, indef- 
initely. Extend one of the sides of the pan, as, for 
example, D 0, until it meets the center line in the 
point B. Still greater accuracy will be insured "by ex- 
tending the opposite side of the pan also, as shown — 
the three lines meeting in the point B — which deter- 
mines the apex of the cone to a certainty. Then B 
and B D, respectively, are the radii of the arcs which 
contain the pattern. From B or any other convenient 
point as center, with B as radius, strike the arc P 
Q indefinitely, and likewise from the same center, 
with B D as radius, strike the arc E F indef- 
initely. From the center B draw a line across these 
arcs near one end, as P E, which will be an end 
of the pattern. By inspection and measurement of the 
plan determine in how many pieces the pan is to be 
constructed and divide the circumference of the pan 
into a corresponding number of equal parts, in this case 
three, as shown by K, M and L. With the dividers or 
spacers step off the length of one of these parts, as 
shown from M to L, and set off a corresponding num- 
ber of spaces on the arc E F, as shown. Through the 
last division draw a line across the arcs toward the 
center B, as shown by F Q. Then P Q F E will be 
the pattern of one of the sections of the pan, as shown 
in the plan. 




Fig. 451. — The Envelope of the Frustum of a Right Cone. 



PROBLEM 124. 



To Construct a Ball in any Number of Pieces, of the Shape of Zones. 



In Fig. 452, let A I G H be the elevation of a 
ball which it is required to construct in thirteen pieces. 
Divide the profile into the required sections, as shown 
by 0, 1, 2, 3, 4, etc., and through the points thus ob- 
tained draw parallel horizontal lines, as shown. The 
divisions in the profile are to be obtained by the fol- 
lowing general rule, applicable in all such cases : 
Divide the whole circumference of the ball into a num- 
ber of parts equal to two times one less than the num- 
ber of pieces of which it is to be composed. 

In convenient proximity to the elevation, the 
center being located in the same vertical line A 1ST, 
draw a plan of the ball, as shown by K M L 1ST. Draw 
the diameter K. L parallel to the lines of division in the 
elevation. With the J-square placed at right angles 



to this diameter, and brought successively against the 
points in the elevation, drop corresponding points upon 
it, as shown by 1, 2, 3, 4, etc. Through each of these 
points describe circles from the center by which the 
plan is drawn. Each of these circles becomes the 
plan of one edge of the belt in the elevation to which 
it corresponds in number, and is to be used in estab- 
lishing the length of the arc forming the pattern of 
the zone of which it is the base. Extend the center 
line NA in the direction of indefinitely. Draw 
chords to the several arcs into which the profile ha3 
been divided, which produce until they cut G A 0, as 
shown by 1 2 E, 2 3 D, 3 4 C, 4 5 B and 5 6 A. Then 
E 2 and E 1 are the radii of parallel arcs which will 
describe the pattern of the first division above the cen- 



252 



Tlie New Metal Worker Pattern Book. 



ter zone, and D 3 and D 2 are the radii describing the 
pattern of the second zone, and so on. 

From E' in Fig. 453 as center, with E 2 and E 1 
as radii, strike the arcs 2 2 and 1 1 indefinitely. Step 



the centers D', Fig. 454; C, Fig. 455; B', Fig. 456, 
and A 1 , Fig. 457. The pattern for the smallest section, 
as indicated by F in the plan, may be struck by a 
radius equal to F 6 in the plan. The center belt or 





"Pig. 453— Pattern of Zone 1 2. 





Fig. 455.— Pattern of Zone 3 4. 




Fig. 456— Pattern of Zone 4 5, 




Fig. 457.— Pattern of Zoup R * 



Fig. 452.— Plan and Elevation 



Fig. 454.— Pattern of Zone 2 3. Fig. 458.— Pattern of Middle Zone. 

To Construct a Ball in any Number of Zones. 



off the length on the corresponding plan line, and make 
1 1 equal to the whole of it, or a part, as may be de- 
sired — in this case a half. In like manner describe 
patterns for the other pieces, as shown, struck from 



zone, shown in the profile by 1 0, is a flat band, and 
is therefore bounded by straight parallel lines, the 
width being 1 in the elevation, and the length meas- 
ured upon line 1 of the plan, all as shown in Fig. 458. 



Pattern Problems. 

PROBLEM 125. 
The Patterns for a Semicircular Pipe with Longitudinal Seams. 



253 



By the nature of the problem the pipe resolves it- 
self, with respect to its section, or profile, into some 
regular polygon. In the illustration presented in Fig. 
4-59 an octagonal form is employed, but any other reg- 



the profile ABCDFHGE and project the points 
B and C back upon N R and complete the elevation by 
drawing the semicircles U and P T. 

By inspection of the diagram it is svident that the 
pattern for the sections corresponding 
to O U T P in the elevation may be 
pricked directly from the drawing as it 
is now constructed, and that the pat- 
terms for the sections represented by B 
A and D F of the profile will be plain 
straight strips of the width of one side 
of the figure, as shown by either E A or 
D F, and in length corresponding to the 
length of the sweep of the elevation on 
the lines NLV and RXS, respectively. 
By virtue of the bevel or flare of 
the pieces N L V U T and R X S T 
P, as shown by A B and C D of 




Ft£. 459.— Elevation and Section 



Fig. 460.— Pattern. 



A Semicircular Pipe with Longitudinal Seams. 



ular shape may be used, and the patterns for it will be 
cut by the same rule as here explained. Let NLYbe 
any semicircle around which an octagonal pipe is to be 
carried. Draw N V, passing through the center W. 
Through W draw the perpendicular L K indefinitely. 
Let N R be the required diameter of the octagon. 
Immediately below and in line with N R construct 



the profile, each becomes one-half of the frustum of a 
right cone, with its apex above or below the point W. 
Therefore prolong C D of the profile until it cuts the 
center line L K of the elevation in the point M. Then 
M D and M G are the radii of the pieces corresponding 
to P T S R of the elevation. Also prolong the side A B, 
or, for greater convenience, its equivalent, E G, until it 



254 



The New Metal Worker Pattern Book. 



cuts the center line in the point M 1 . Then M' G and 
M 1 E are the radii of the pieces corresponding toNLV 
U of the elevation. From M 2 in Fig. 460 as center, 
using each of the several radii in turn, strike arcs in- 
definitely, as shown by N' V, O 1 U\ P 1 T 1 and B* S 1 . 
Step off the length NLYin the elevation, Fig. 457, 
and make N 1 V 1 of Fig. 458 equal to it. Draw N' O 1 
and V U' radial to M\ Then N 1 V IP O 1 will con- 
stitute the pattern for the pieces N L V U of the 
elevation. In like manner establish the length of P 1 
T', and draw P 1 R 1 and T 1 S 1 , also radial to the center, 
as shown. Then P' T' S 1 R 1 will be the pattern for 
the pieces P T S X R of the elevation. 

This rule may be employed for carrying any polyg- 
onal shape around any curve which is the segment of a 



circle. The essential points to be observed are the 
placing of the profile in correct relationship to the ele- 
vation and to the central line L K, after which prolong 
the oblique sides until they cut the central line, thus 
establishing the radii by which their patterns may be 
struck. In the case of elliptical curves, by resolving 
them into segments of circles and applying this rule 
to each segment, as though it were to be constructed 
alone and distinct from the others, no difficulty will be 
met in describing patterns by the principles here set 
forth. The several sections may be united so as to 
produce a pattern in one piece by joining them upon 
their radial lines. This principle is further explained 
in the pattern for the curved molding in an elliptical 
window cap in Problem 128. 



PROBLEM 126. 

The Blank for a Curved Molding:. 



As curved moldings necessitate a stretching of the 
metal in order to accommodate them to both the curve 
oi the elevation or plan and the curve of the profile at 



machinery designed for that purpose, care being 
taken to make the width of the flaring strip suffi- 
cient to include the stretchout of the curve of the pro- 



M,\ s 





Fig. 461. — Obtaining the Blank for a Curved 
Cove or Ovolo Molding. 



Fig. 462.— Obtaining the Blank for a Curved 
Ogee Molding. 



the same time, the patterns for their blanks can only 
be considered as flaring strips of metal in which the 
curve of the elevation or plan only is considered. The 
curve of the profile requires to be forced into them by 



file. Blanks for curved moldings thus become frus- 
tums of cones and are cut according to the principles 
of regular flaring articles, as explained in the preceding 
problems. The method of determining the exact flare 



Pattern Problems. 



255 



necessary to produce a certain mold with the greatest 
facility is a matter to be determined by the nature of 
the profile and the kind of machinery to be used in 
forming the same. Usually a line is drawn through 
the extremities of the profile, as shown at A D in 
either of the two illustrations here given, Figs. 461 
and 462, and is continued until it meets the center line, 
for length of radius, as shown at F. 

Therefore, to describe the pattern of the blank 
from which to make a curved molding corresponding 
to the elevation A C E D, proceed in the same manner 



as though the side E C were to be straight. Through 
the center of the article draw the line B F indefinitely, 
and draw a line through the points C and E of one of 
the sides, which produce until it meets B F in the 
point F. Then F E will be the radius of the inside of 
the pattern. The radius of the outside is to be ob- 
tained by increasing F G an amount equal to the excess 
of the curved line E C over the straight line E C, as 
shown by the distance C S. Then F S is the radius of 
the outside of the pattern. The length of the pattern 
can be obtained as in previous problems. 



PROBLEM 127. 
The Patterns for Simple Curved Moldings in a Window Cap. 



In Fig. 463 is shown the elevation of a window 
cap, in the construction of which two curved moldings 
are required of the same profile, but curved in opposite 



essary for joining it to the face and roof pieces will be 
obtained in one piece. The method of developing the 
pattern for the blank is the same for both curves. The 




Fig. 463.— Elevation of Window Cap. 




Fig. 464.— Blank for Center Pieoe. 



e 


1 






/-\_ 


1 


jL_ 


8 




*■--—-. "* 9 




Fig. 465.— Blank for Side Piece. 



The Patterns for Simple Curved Moldings in a Window Cap. 



directions. It is advisable to include as much in one 
piece as can be raised conveniently with the means at 
hand ; therefore, the curved part of the profile with its 
fillets or straight parts adjacent and the two edges nec- 



two pieces will raise to the form by the same dies or 
rolls, it being necessary only to reverse them in the 
machine. Before the blank for the middle piece can 
be developed it will be necessary to first construct a 



256 



The New Metal Worker Pattern Book. 



section upon the center line, as shown at S K ; from all 
points in the mold and the center of the curve upon the 
center line project horizontal lines to the right. Draw 
any vertical line, as H K, to represent the face of the 
cap in the section and at S draw the profile of the 
mold, as shown. The principle to be employed in strik- 
ing the pattern is simply that which would be used in 
obtaining the envelope of the frustum of a cone of 
which A D is the axis. 

The general average of the profile is to be taken 
in establishing the taper of the cone, or, in other 
words, a line is passed through its extreme points. 
Draw a line through the profile in this manner and pro- 
long it until it intersects A D in the point A, all as 
shown by C A. Then A is the apex of the cone, of 
which A C is the side and H D the top of the frustum. 
Divide the profile S, as in ordinary practice for stretch- 
outs, into any number of spaces, all as shown by the 
small figures. Transfer the stretchout of the profile S 
on to the line A C, commencing at the point 1, as 
shown, letting the extra width extend in the direction 
of C. From any convenient center, as A in Fig. 464, 
with radius A C, describe the pattern, making the 
length of the arc equal to the length of the correspond- 
ing arc in the elevation, all as shown by the spaces and 
numbers. From the same center draw arcs correspond- 



ing to points 9, 10 and 11 of the stretchout, thus com- 
pleting this pattern. 

For the pattern of the curved molding forming the 
end portion of the cap proceed in the same general 
manner. Upon any line drawn through the center N 
of the curve, as L M, construct a section of the mold, 
as shown at R. From N draw the perpendicular N B 
indefinitely. Through the average of the profile R, as 
before explained, draw the line to B, cutting N Bin 
the point B, as shown. Lay off the stretchout of the 
profile upon this line, commencing at the point 1, in 
the same manner as explained in the previous operation. 
From any convenient point, as B' in Fig. 465, as center, 
with radius B 1, describe the inner curve of the pat- 
tern, as shown, which in length make equal to the eleva- 
tion, measuring upon the arc 1, all as shown by the 
small figures, after which add the outer curves, as 
shown by E 1 E\ 

The straight portion forming the end of this mold- 
ing, as shown in the elevation, is added by drawing, at 
right angles to the line E 5 B 1 , a continuation of the 
lines of the molding of the required length, as shown 
in the pattern. Upon this end of the pattern a 
square miter is to be cut by the ordinary rule for 
such purposes, to join to the return at the end of the 
cap. 



PROBLEM 128. 

The Pattern for the Curved Molding in an Elliptical Window Cap. 



In Fig. 466 is shown the elevation and vertical sec- 
tion of a window cap elliptical in shape, the face of which 
is molded. In drawing the elevation such centers have 
been employed as will produce the nearest approach to 
a true ellipse after the manner described in Problem 76 
of Geometrical Problems, page 65. The centers B, D and 
F, from which the respective segments of the elevation 
have been described, may then be used in obtaining pat- 
terns as follows : Through the center F, from which 
the arc forming the middle part of the cap is drawn, 
and at right angles to the center line of the cap Gr H, draw 
the line I K indefinitely. Project a section on the center 
line of the cap, as shown by P K at the right, the line 
P K being used as a common basis of measurement upon 
which to set off the semi-diameters of the various cones 
of which the blanks for the moldings form a part. 
Through the average of the profile, as indicated, draw 
S R, producing the line until it meets I K. Divide the 
profile of the molding in the usual manner and lay off 



the stretchout, as indicated by the small figures. Then 
R S is the radius of the pattern of the middle segment 
of the cap. 

With the dividers, measuring down from the pro- 
file, lay off on P K distances equal to the length of 
the radius A B, as shown by the point 0, and of C D, 
as shown by the point M. Through these points 
and M, at right angles to P K, draw lines cutting S R 
in the points T and U. Then U S is the radius for 
the pattern of the segment C E of the elevation, and T S 
the radius of the pattern for the segment A C. In 
order to obtain the correct length of the pattern, not 
only as regards the whole piece, but also as regards 
the length of each arc constituting the curve, step off 
the length of the curved molding with the dividers 
upon any line of the elevation most convenient, as 
shown, numbering the spaces as indicated, and setting 
off a like number of spaces upon a corresponding line 
of the pattern. As a matter both of convenience and 



Pattern Problems. 



257 



accuracy, the spaces used in measuring the arcs are 
greater in the one of longest radius and are diminished 
in those of shorter radii, as will be noticed by examina- 
tion of the diagram. 

To lay off the pattern after the radii are obtained 
as above described, proceed as follows : Draw any 
straight line, as G' H 1 in Fig. 467, from any point 
in which, as F', with radius equal to R S, as shown 
by F 1 E 1 , describe an arc, as shown by E 1 G 1 ; and 
likewise, from the same center, describe other arcs cor- 
responding to other points in the stretchout of the pro- 
file. Make the length of the arc E' G 1 equal to the 
length of the corresponding arc in the elevation, as de- 
scribed above. From E 1 to the center F', by which this 
arc was struck, draw E 1 F 1 . Set the dividers to the 
distance U S as radius, with which, measuring from E 



length of the corresponding arc in the elevation, 
all as shown by the small figures. From C draw the 
line C 1 D 1 to the center by which this arc was struck. 




Fig. 467.— Blank for the 
Curved Molding. 



Fig. 466.— Elevation and Section of Window Cap.' 

The Pattern for the Curved Molding in an Elliptical Window Cap. 



along the line E 1 F", establish D' as center, from which 
describe arcs corresponding to the points in the profile, 
as shown from E 1 to C. Make E 1 C equal to the 



Set the dividers to the distance T S in the section, 
and, measuring from C along the line C D 1 , establish 
the point B 1 , from which as a center strike arcs cor- 



258 



Hie New Metal Worker Pattern Book. 



responding to those already described in the other 
section of the pattern. Make the length equal to the 
length of the corresponding segments in the elevation, 



and draw the line A 1 B 1 . Then A 1 C E 1 G l is 
the half pattern corresponding to A C E G of the 
elevation. 



PROBLEM 129. 



The Pattern of an Oblong: Raised Cover with Semicircular Ends. 



In Fig. 468 let A B C D represent a side eleva- 
tion of the cover of which E G F H is the plan or 
shape of the vessel it is to fit. Various constructions 
may be employed in making such a cover as this; 
that is, the joints, at the option of the mechanic, may 
be placed at other points than shown here ; the prin- 
ciple used in obtaining the shape, however, is the 
same, whatever may be the location of the joints. 
By inspection of the elevation and plan it will be seen 
that the shape consists of the two halves of the en- 
velope of a right cone, joined by a staiight piece. 
Therefore, for the pattern proceed as follows : At any 
convenient point lay off B 2 C 2 , in length equal to B 1 
C 1 of the plan. From B 2 and C 2 as centers, with 
radius equal to A B or C D of the elevation, describe 
arcs, as shown by N and P M. Upon these arcs, 
measured from and P, respectively, set off the 
stretchout of the semicircular ends, as shown in plan, 
thus obtaining the points M and N. From N draw 
N B', and from M draw M C\ From B 2 and C 2 , at 
right angles to the line B 2 C 2 , draw B 2 K and C 3 L, in 
length equal to A B of the elevation, which represents 



e 


C 


^^ 




^\ 












G 






! V^ 






~^~\ j 


Y 






\| 


E'\ B ' 






J 






V 








' Xn^ 






^y 


• ■ 


H 






6 


P 




i^y* 








-~f 








J 2 
B 






lc* 










' N. / 








-v 









Fig. 468. — Elevation, Plan and Pattern of an Oblong Raised Cover 
with Semicircular Ends. 



the slant hight of the article. Connect K and L, as 
shown. Then N K L M P will be the required 
pattern. 



PROBLEM 130. 



The Pattern of a Regular Flaring- Article which Is Oblong- with Semicircular Ends. 



In Fig. 469, let A B D C be the side elevation of 
the required article. Below it and in line with it draw 
a plan, as shown by E c d F H G. From D in the 
elevation erect the perpendicular D L. Then L C 
represents the flare of the article and C D is the width 
of the pattern throughout. Across the plan, at the 
point where the curved end joins the straight sides, 
draw the line d H at right angles to the sides of the 
article. As the plan may be drawn at any distance 



from the elevation, this line must be prolonged, if 
necessary, to meet C D extended. Produce C D until 
it meets d H, as shown by g. Then g D and g C are 
radii of the curved parts of the pattern. 

Lay off on a straight line M in Fig. 470, the 
length of the straight part of the article, as shown in 
the plan by c d. At right angles to M draw M S 
and B, indefinitely. Upon these lines set off from 
M and the distance g C, locating the points S and R, 



Pattern Problems. 



259 



the centers for the curved portions of the pattern. 
From S with the radius g C strike the arc M U indef- 
initely. In like manner, with same radius, from E as 
center describe the arc V. From the same centers, 
with radius equal to g D, describe the arcs N T and P 
"W. Step off the length of the curved part of the article 
upon either the inner or outer line of the plan, and make 
the corresponding arc of the pattern equal to it, as 
shown by the spaces in N T and P W. Through the 
points T and "W draw lines from the centers S and E, 
producing them until they cut the outer arcs at IT and 
V. At right angles to the line STUorEW V, as 
the case may be, set off V X Y W, equal to M P N, 
which will be the other straight side of the pattern. 
Then U M V X Y W P N T will be the complete 
pattern in one piece. 

If it were desired to locate the seam midway in 



end, instead of all at one end, as shown. In like man- 
ner such changes may be made as are necessarv for 





Fig. 469.— Plan and Elevation. 



Pig-. 470.— Pattern. 



The Pattern of a Regular Flaring Article Which is Oblong with Semicircular Ends. 



one of the straight sections, in adding the last member 
as above described, one-half would be placed at each 



locating the seam at any other point, or for cutting the 
pattern in as many pieces as desired. 



PROBLEM 131. 



The Pattern of a Regular Flaring Oblong Article with Round Corners. 



In Fig. 471, A C D B is the side elevation of the 
article and EFGMNOPEthe plan. The corners 
are arcs of circles, being struck by centers H, L, T 
and S, as shown. Draw the plan in line with the ele- 
vation, so that the same parts in the different views 



shall correspond. Through the centers H and L of 
the plan by which the corners F Gr and M N are struck, 
draw F N indefinitely. Prolong the side line of the 
elevation C D until it cuts F 1ST in the point K, as 
shown. Then K D is the radius of the inside line of 



260 



TJie New Metal Worker Pattern Book. 



the pattern of the curved part, and K C is the radius 
of the outside line. 

Draw the straight line E 1 F of Fig. 472, in length 
equal to the straight part of one side of the article, or E 
F of the plan. Through the points E 1 and F 1 , at right 
angles to the line E' F 1 , draw lines indefinitely, as shown 
by E 1 U and F 1 K 1 . Upon these lines set off, from F 1 and 
E', the distance K C, locating the points K 1 and U, the 
centers for the curved parts. From K 1 , with the radius 
K C, strike the arc F 1 G 1 , which in length make equal 
to F G of the plan. From G 1 draw a line to the center 
K', at right angles to which erect G' M 1 , in length 
equal to G M of the plan. In like manner, with like 
radius, describe the arc E 1 R 1 . Draw R 1 U, at right 




F / 



«U 



N 
Fig. 471.— Plan and Elevation. 



right angles to it lay of! 0' N 1 , equal to N of the 
elevation. Draw 1ST 1 W, and draw the arc N 1 M* in the 




Fig. 473.— Pattern. 



The Pattern of a Regular Flaring Oblong Article With Round Corners. 



angles to which erect R 1 P 1 , equal to R P of the plan. 
At right angles to R' P 1 draw P 1 V indefinitely. In 
the manner above described establish the center V, and 
from it describe the third arc P l 0'. Draw 1 V. At 



same manner as already described. In the same man- 
ner lay off the inner line of the pattern, as shown by 
mgferpon m l . Join the ends M 1 m and M 3 m 1 , thus 
completing the pattern sought. 



PROBLEM 132. 

The Envelope of the Frustum of a Cone, the Base of Which Is an Elliptical Figure. 



This shape is very frequently used in pans and 
plates, and therefore in Fig. 473 is shown an elevation 
and plan of what is familiarly termed an oval flaring 



pan. Let that part of the plan lying between H and 
L be an arc whose center is at U, and let those por- 
tions between V and H and L and W be arcs whose 



Pattern Problems . 



261 



centers are, respectively, R and S. A C D B represents 
an elevation of the vessel, and is so connected with the 
plan as to show the relationship of corresponding 
points. 

The first step is to construct a diagram, shown in 
Tig. 474, by means of which the lengths of the radii to 
be used in describing the pattern are to be obtained. 
Draw the horizontal line H U indefinitely, and at right 
angles to it draw H A, indefinitely also. Make II U 
equal to H U of the plan, Fig. 473. Make H C equal 
to the vertical hight of the vessel, as shown in the 




N 0. From the points H and L of the arc first drawn 
draw lines to A, thus intercepting the arc N and 
determining its length. 

In the diagram, Fig. 474, set off from H, on the 
line H U, the distance R H, making it equal to E H 
of the plan, Fig. 473. Also, upon the line C G, from 
the point C, set off C I, equal to E N of the plan. 
Then, through the points R and I thus established, 
draw the line R B, which produce until it intersects 
A II. Then R B will be the radius for those portions 
of the pattern lying between V and H and L and W of 




Fig. 473.— Plan and Elevation. 



R U 

Fig. 474.— Diagram of Radii. 




Fig. 475.— Pattern of One Half. 



The Envelope of the Frustum of a Cone, the Base of Which is an Elliptical Figure. 



elevation by D X. Draw the line C G parallel to H U, 
making C G in length equal to U N of the plan. 
Through the points U and G thus established draw the 
line U G, which continue until it meets H A in the 
point A. Then A U will be the radius by which to 
describe that portion of the pattern which is included 
between the points H and L of the plan. With A U 
as radius, and from any convenient point as center — as 
A, Fig. 475 — draw the arc H L, which in length make 
equal to H L of the plan, Fig. 473, as shown by the 
points 1, 2, 3, etc. From the same center, and with 
f,he radius A G of Fig. 474, describe the parallel arc 



the plan. From the point H, on the line H A, Fig. 
475, set off the distance H B, equal to R B of Fig. 
474. Then, with B as center, describe the arc E H, 
and from a corresponding center, C, at the opposite end 
on pattern, describe the arc L K. From the same 
centers, with B I as radius, describe the arcs N M and 
P, all as shown. Make H E and L K in length 
equal to H E and L K of the plan. From E and K, 
respectively, draw lines to the centers B and C, inter- 
cepting the arcs 1ST M and P in the points M and P. 
Then E K P M will be one-half of the complete pat- 
terns of the vessel. 



PROBLEM 133. 

The Pattern of a Heart-Shaped Flaring: Tray. 



Let E C G 1 F G C of Fig. 476 be the plan of the 
article, and INOK the elevation. By inspection of 
the plan it will be seen that each half of it consists of 



two arcs, one being struck from D or D 1 as center, and 
the other from C or C as center, the junction between 
the two arcs being at G and G 1 , respectively. From G 



262 



The New Metal Worker Pattern Book. 



draw C F, and likewise draw C G. Upon the point 
D : erect the perpendicular D 1 C 1 . 

To obtain the radii of the pattern construct a dia- 



Lay off the perpendiculars X U and P S indefinitely. 
Upon P S, from P, set off P E, equal to D 1 C of the 
plan, and on X U, from X, set off X "W, equal to D 1 c of 




Pig. *76.— Plan and iteration. 



Fig. 178— Pattern. 



Fig. 477.— Diagram of Kadi! 



The Pattern of a Heart-Shaped Flarin? Tray. 



gram, shown in Fig. 477, which is in reality a section 
upon the line C G of the plan. Draw X P in Fig. 477, 
in length equal to the straight hight of the article. 



the plan. In like manner make P S equal to C G of 
the plan, and X U equal to C g of the plan. Connect 
U S and W R. Produce P X indefinitely in the direc- 



Pattern Problems. 



263 



tion of Z. Also produce E "W until it meets P X in 
the point Y, and in like manner produce S U until it 
meets P Z in the point Z. Then Z U and Z S are the 
radii for that portion of the article contained between 
G and F of the plan, and Y "W and Y E are the radii of 
that portion shown from G to E of the plan. 

To lay out the pattern after the radii are estab- 
lished, draw any straight line, as Z" G 2 in Fig. 478, in 
length equal to Z S of the diagram. From Z 1 as cen- 
ter, with Z S as radius, describe the arc G 2 F 1 , in length 



equal to G F of the plan. In like manner, with radius 
Z U, from the same center, describe the arc g 1 /', in 
length equal to g/ol the elevation. Draw/ 1 F 1 . Set 
off from G 2 , upon the line G 2 Z", the distance E Y of 
Fig. 477, as shown at Y 1 , and from Y 1 as center, with 
the radius E Y, describe the arc G 2 E 1 , which in length 
make equal to G E of the plan. In like manner, from 
the same center, with radius Y ~W, describe the arc 
g l e\ equal to the arc g e of the plan. Draw e' E 1 , thus 
completing the required pattern. 



PROBLEM 134. 



The Pattern of an Oval or Egg-Shaped Flaring Pan. 



Let A B C D in Fig. 479 represent the elevation 
of the article, of which A 1 K L B 1 M I is the plan. 
The plan is constructed by means of the centers 0, P, 
F and F 1 , as indicated. The patterns, therefore, are 



construct a section of the article as it would appear if 
cut on the line A 1 P of the plan. Therefore set off, 
at right angles to it, A 2 P 2 , equal to A 1 P. Make 
P 2 D 3 equal to the straight bight of the article, as 



At-— 1— ;P 




Fig. 480.— Diagram of Small Cone. 



&-- B' 





k 



Fig. 479.— Plan and Elevation. 



Fig. 481.— Diagram of Middle Cone. 
An Oval or Egg-Shaped Flaring Pan. 



Fig. 482.— Diagram of Large Cone 



struck by radii obtained from sections of the several 
cones of which the article is composed. At any con- 
venient place draw the line P 2 P 1 , Fig. 480, indefi- 
nitely, corresponding to P of the plan, and upon it 



shown by E D of the elevation. Make D 2 A 3 of the 
diagram equal to D 1 P of the plan. Draw A 2 A 3 , 
which will correspond to A D of the elevation, pro- 
longing it until it meets P 2 P 1 in the point P 1 . Then 



264 



T/ie New Metal Worker Pattern Book. 



P' A 3 is the radius of the outside line of the pattern 
of the portion between K and I of the plan, and P 1 A 3 
is the radius of the line inside of the same part. 

In like manner draw the line 1 2 , Fig. 481, 
corresponding to of the plan, and construct a section 
taken on the line B 1 , as shown by 2 B 2 C 2 C 3 . 
Produce B" C 2 until it meets 2 1 in the point 1 . 
Then 1 C 2 and O 1 B 2 are the radii of the pattern of 
that portion of the article contained between L and M 
of the plan. 

Draw the line F 3 F 2 , Fig. 482, which shall corre- 
spond to F or F 1 of the plan. Make F 3 E equal to the 
straight hight of the article, and lay off F' L 3 at right 
angles to it, equal to F 1 L of the plan, and E F equal 
to F 1 I of the plan. Draw L 9 Z 2 , which produce until 
it meets F 3 F 2 in the point F 3 . Then F 2 I and F 2 L 3 , 
respectively, are the radii of the pattern of those parts 
shown by K L and I M of the plan. 

To lay off the pattern after the several radii are 
obtained, as described above, draw any straight line, 
in length equal to F 2 L 3 , as shown by F 4 K in Fig. 
483, and from F 4 as center, with F 2 P and F 2 L 3 , Fig. 
482, as radii, strike arcs, as shown by k' V and K L 1 , 
which in length make equal to the corresponding arcs 
of the plan K L and k I, as shown. Draw L 1 F 4 , and 
upon it set off from t toward F 4 , a distance equal to 
O' C 2 of Fig. 481, establishing the point 3 , from 
which strike the arc T m\ in length equal to I m of 
the plan. In like manner, from the same center, 
with radius 0' B 2 , of Fig. 481, strike the arc L' M 1 , 
equal in length to L M of the plan. Draw M' 3 , 
which produce in the direction of F' making M' F b 
equal to K F 4 , from which center continue the inner 
line of the pattern, as shown by m 1 i\ which in length 
must equal m i of the plan. In like manner, from the 
same center, with radius F 3 L 3 of Fig 4S2, describe 
the arc M 1 1 1 and draw I 1 F 5 . Set off on this line from 
i' a distance equal to P 1 A 3 of Fig. 480, thus estab- 
lishing the center P 3 . Describe the arc i l k, in length 
equal to i k of the plan. In like manner, from the 
same center, with the radius P 1 A 3 of Fig. 480 describe 
the arc I' K\ in the length equal to I K of the plan. 
Place the straight edge against the points P 3 and K 2 
and draw K 3 k, thus completing the pattern. 



From inspection it is evident that the pattern 
might have been commenced at any other point as 
well as at K k of the plan. If the joint is desired 
upon any of the other divisions between the arcs, as 
L /, M m, or I i, the method of obtaining it will be so 
nearly the same as above narrated as not to require 
special description. If the joint is wanted at some 
point in one of the arcs of the plan, as, for example, 




Fig. 488. — The Pattern of an Oval or Egg-Shaped Flaring Pan. 

at X x, draw the line X x across the plan, producing 
it until it meets the center by which that arc of the 
plan is struck. In laying off the pattern, commence 
with a line corresponding to X F', in place of F 4 K 1 , 
and from it lay off an arc corresponding to the portion 
of the arc in the plan intercepted by X x, as shown by 
XL';c. Proceed in other respects the same as above 
described until the line k K 2 is obtained, against which 
there must be added an arc corresponding to the 
amount cut from the first part of the plan by X x, as 
above described, or, in other words, equal to X K k x 
of the plan. 



Pattern Problems. 

PROBLEM 135. 
The Envelope of the Frustum of a Right Cone, the Upper Plane of Which Is Oblique to Its Axis. 



265 



In Fig. 484, let C B D E be the elevation of the 
required shape. Produce the sides C B and E D un- 
til they intersect at A. Then A will be the apex of 




Fig. 434- — The Envelope of the Frustum of a Right Cone Whose Upper Plane is 

Oblique to its Aotis. 



the cone of which C B D E is a frustum. Draw the 
axis A G, which produce below the figure, and from a 



center lying in it draw a half plan of the article, as 

shown by FGH. 

Divide this plan into any number of equal parts, 
and from the points carry lines 
parallel to the axis until they cut 
the base line, and from there extend 
them in the direction of the apex 
until they cut the upper plane B D. 
Place the T-square at right angles 
to the axis, and, bringing it against 
the several points in the line B D, 
cut the side A E, as shown. From 
A as center, with A E as radius, 
describe the arc C 1 E', on which 
lay off a stretchout of either a half 
or the whole of the plan, as may 
be desired, in this case a half, as 
shown. From the extremities of 
this stretchout, C and E', draw lines 
to the center, as C A and E 1 A. 
Through the several points in the 
stretchout draw similar lines to the 
center A, as shown. With the 
point of the compasses set at A, 
bring the pencil to the point D in 
the side A E, and with that radius 
describe an arc, which produce until 
it cuts the corresponding line in the 
stretchout, as shown at D 1 . In like 
manner, bringing the pencil against, 
the several points between D and E. 
in the elevation, describe arcs cut- 
ting the corresponding measuring 
lines of the stretchout. Then a line 
traced through these intersections 
will form the upper line of the 

pattern, the pattern of the entire half being con- 

tained in 0" B 1 D 1 E 1 . 



PROBLEM 136. 

The Envelope of a Right Cone Whose Base Is Oblique to Its Axis. 

In Fig. 485, let G D H be the elevation of a right I which is required. It will be necessary first to assum- 
cone whose base is oblique to its axis, the pattern of ( any section of the cone at right angles to its axis as a 



266 



TJie New Metal Worker Pattern Book. 



base upon which to measure its circumference. This 
can be taken at any point above or below the oblique 
base according to convenience. 

Therefore at right angles to the axis D 0, and 
through the point G, draw the line G F. Extend the 
axis, as shown by D B, and upon it draw a plan of 
the cone as it would appear when cut upon the line G F, 
as shown by ABC. Divide the plan into any convenient 
number of equal parts, and from the points thus ob- 
tained drop lines on to G F. From the apex D, 
through the points in G F, draw lines to the base 
G II. From D as center, with D G as radius, describe 
an arc indefinitely, on which lay off a stretchout taken 
from the plan ABC, all as shown by I M K. From 
the center D, by which the arc was struck, through 
the points in the stretchout, draw radial lines indefi- 
nitely, as shown. Place the blade of the T-square 
parallel to the line G F, and, bringing it against the 
several points in the base line, cut the side D H, as 
shown, from F to H. With one point of the com- 
passes in D, bring the other successively to the points 
1, 2, 3, i, etc., iu F H, and describe arcs, which pro- 
duce until they cut the corresponding lines drawn 
through the stretchout, as indicated by the dotted 
lines. Then a line, I L K, traced through these 
points of intersection, as shown, will complete the re- 
quired pattern. 



Pattern 




Elevation 



Fig. 485. — The Envelope of a Cone whose Base is Oblique to its Axis 



PROBLEM 137. 

A Conical Flange to Fit Around a Pipe and Against a Roof of One Inclination. 



In Fig. 486 is shown, by means of elevation and 
plan, the general requirements of the problem. A B 
represents the pitch of the roof, G H K I represents the 
pipe passing through it, and C D F E the required flange 
fitting around the pipe at the line C D and against 
the roof at the line E F. The flange, as thus drawn, 
becomes a portion of the envelope of a right cone. 

At any convenient distance below the elevation 
assume a horizontal line as a base of the cone upon 
whioh to measure its diameter, and continue the sides 
downward till they intersect this base line, all as shown 
at L M. Also continue the sides upward till they in- 
tersect at W, the apex. Below the elevation is shown 
a plan, and similar points in both views are connected 
by the lines of projection. S T represents the pipe 
and N the flange. While the pipe is made to pass 
through the center of the cone, as may be seen by ex- 
amining the base line L M in the elevation, and also 



P R of the plan, it does not pass through the center of 
the oblique cut E F in the elevation, or, what is the 
same, N of the plan. 

For the pattern of the flange proceed as shown in 
Fig. 4S7, which in the lettering of its parts is made to 
correspond with Fig. 4S6, just described. Divide the 
half plan PXE into any convenient number of parts — in 
this case twelve — and from each of the points thus 
established erect perpendiculars to the base of the cone, 
obtaining the points 1', 2 1 , 3 1 , etc. From these points 
draw lines to the apex of the cone W, cutting the 
oblique line E F and the top of the flange C D, as 
shown. Inasmuch as C D cuts the cone at right angles 
to its axis, the line in the pattern corresponding to it 
will be an arc of a circle ; but with E F, which cuts the 
cone obliquely to its axis, the case is different, each 
point in it being at a different distance from the apex. 
Accordingly, the several points in E F, obtained by 



Pattern Problems. 



267 



the lines from the plan drawn to the apex W, must be 
transferred to one of the sides of the cone, where their 
distances from W can be accurately measured. There- 
fore from the points 3 , l 3 , 2 3 , 3 3 , in E F, draw lines 
at right angles to the axis of the cone W X, cutting 
the side W M, as shown. With W as center, and with 
W M as radius, strike the arc P' B 1 indefinitely, and, 



in the plan P X B, all as shown by 2 , P, 2 2 , 3", etc. 
From these points draw lines to the center W, as shown. 
With one point of the dividers set at W and the other 
brought successively to the points obtained in W M by 
the horizontal lines drawn from E F, cut the correspond- 
ing lines in the stretchout of the pattern, as indicated 
by the curved dotted lines. A line traced through 




Fig. 4S6.— Plnn and Elevation. 




. 487.— Pattern. 



A Conical Flange to Fit Around a Pipe and Against a Roof of One Inclination. 



with the same center and with W D as radius, strike 
the arc C D 1 indefinitely, which will form the bound- 
ary of the pattern at the top. At any convenient dis- 
tance from M draw W P 1 , a portion of the length of 
which will form the boundary of one end of the pat- 
tern. On P 1 E', commencing with P 1 , set off spaces 
equal in length and the same in number as the divisions 



these points, as E 1 F', will represent the lower side of 
the pattern. As but one-half of the plan has been 
used in laying out the stretchout, the pattern C E'_ F 1 
D 1 thus obtained is but one-half of the piece required. 
It can be doubled so that the seam can be made to 
come through the short side at C E, or through the 
long side at D F, at pleasure. 



PROBLEM 138. 

The Pattern for a Cracker Boat. 



Let E F H Gr, in Fig. 48S, be the side elevation, 
ABODE the end, and I K J L the plan of a dish 
sometimes called a cracker boat or bread tray. The 



sides of the dish are parts of the frustum of a right 
cone. To the plan have been added the circles show- 
ing the complete frustums of which the sides are a part, 



268 



The New Metal Worker Pattern Booh. 



L and K being the centers, all of which will appear 
clear from an inspection of the drawing, and below is 
further shown a side view of this frustum. While in 




pleted cone. Before the pattern can be described it 
will be necessary to draw a half elevation of the cone 
U V Z, showing the end view of the tray in its relation 
to the same, as in Fig. 489. Draw any center line, as 
K' Z'. From the point L', as center, strike the arc K' 
T', being one-fourth of the plan of top, as shown by K 
T in Fig. 48S. Below the plan of top draw one-half 
of frustum of cone, as shown by v! T'X' w', in which 
draw the end elevation of boat A' B' C X' E', letting 
V X' be one of its sides, and extend the line V B' 
through the arc K' T' at B". Divide the part of plan 
B" T" into any convenient number of parts, and from 
the points carry lines parallel to the center line or axis 
until they cut the top line v! V, and from there extend 
them in the direction of Z' until they cut the line B' C 
Place the T-square at right angles to the axis, and, 
bringing it against the several points in the line B' C, 
which represents the shape shown by E C F in eleva- 
tion of side, cut the side V X', as shown. From Z' as 
center, with Z' V as radius, describe the arc I' J', upon 
which lay off a stretchout of plan. As the part of the 
plan B" T' corresponds to B' C, which shows one-half 

of one side of boat, and as 



Fig. 4SS.— Plan and Elevations. Pig-. 489.— Pattern. 

The Pattern for a Cracker Boat. 

the plan the top and bottom of the sides have been 
shown parallel, in the side view the top appears curved 
at C, the cut producing which curve being shown by 
B C of the end view. 

Extend the sides IT W and V X of the frustum 
until they meet at Z, which is the apex of the com- 



this part of plan is divided 
into three parts, six of 
these parts are spaced off 
on the arc I' J and num- 
bered from 1 to 4, and 4 
to 1, 4 being the center 
line. Through these points 
in the stretchout draw 
measuring lines to the cen- 
ter Z', as shown. With 
one point of the compasses 
set at Z', bring the pencil 
point up to the several 
points between V and C in 
the elevation, and describe 
arcs cuttingmeasuringlines 
of corresponding numbers 
in the stretchout; then a line traced through these 
points of intersection will form the line I' K' J', show- 
ing the upper line of the pattern for one side of the 
boat. 

To obtain the bottom line of the pattern, with Z' 
as center and radius Z' X', describe the arc M' X'. 
Divide the plan of bottom of boat, as M T N in Fig. 
48S, into any convenient number of equal parts, in this 
case six, three on each side of the center T, and start- 
ing from the center line 4 of pattern, space off three 
spaces each way on the arc M' N', thus establish- 



Pattern Problems. 



269 



mg rae points M' and 1ST' of pattern, corresponding 
to the points M and X of plan. By drawing the 



lines M' I' and N' J' the pattern for one side of the 
boat, shown by E F H Gr in elevation, is completed. 



PROBLEM 139. 



Pattern for the Frustum of a Cone Fitting Against a Surface of Two Inclinations. 



In Fig. 490, let A B C D represent the frustum 
of a cone, the base of which is to be so cut as to make 
it fit against a roof of two inclinations, as indicated by 
PED. Continue the lines of the sides of the cone 
A B and D C upward until they meet in the point X, 
which is the apex of the complete cone. Through the 
apex of the cone draw the line X B, representing the 
axis of the cone, meeting the ridge of the roof in the 
point R, and continuing downward in the direction of 
Y, as shown. At any convenient distance below A D 
draw a horizontal line, G H, as a 
base, and immediately below it draw 
apian of the same, as shown by E 
SFY. 

Subdivide this plan into any 
convenient number of spaces, as 
indicated by the small figures 0, 1, 
2, 3, etc. From the points thus 
established carry lines vertically 
until they cut the base of the cone 
Gr H, and from this line carry them 
in the direction of the apex X until 
they cut the line of the given roof. 
From the points established in the 
roof line A R draw lines at right 
angles to the axis of the cone X 
Y, continuing them until they strike 
the side of the cone A B. From 
X as center, with X G as radius, describe the arc GrK, 
upon which lay off a stretchout of the plan. 

As the pattern really consists of four equal parts 
or quarters, the divisions of the plan have been num- 
bered from to 4 and from i to alternating, the 
points representing the lowest and the points 4 the 
highest points of each quarter. Therefore in number- 
ing the points of the stretchout Gr K, any point can be 
assumed as a beginning which is deemed the best place 
for the joint (in this case 4), numbering from 4 to and 
reversing each time, all as shown. From these points 
established in the arc G K draw lines to the apex X. 
Then, with X as center, and with radii corresponding to 



the points already established in the side B G of the cone, 
strike arcs as shown by the dotted lines, cutting meas- 




Fig. 490. — Pattern for the Frustum of a Cone Fitting Against a 
Surface of Two Inclinations. 

uring lines of corresponding number. Then a line traced 
through the points of intersection, as shown by L, will 



270 



Tlie New Metal Worker Pattern Book. 



be the shape or the pattern at the bottom and O N M L I cone adapted to set over the ridge of a roof, as indi- 
will constitute the entire pattern of the frustum of a I cated in the elevation. 

PROBLEM 140. 

The Pattern of a Frustum of a Cone Intersected at its Lower End by a Cylinder, Their Axes Intersecting: 

at Right Angles. 



Let S P E T in Fig. 491 be the elevation of the 
cylinder, and a&KHJ the elevation of the frustum. 
Draw the axis of the cylinder, A B, which prolong, as 
shown by C D, on which construct a profile of the 
cylinder, as shown by C E D F. Produce the sides of 
the frustum, as shown in the elevation, until they 
meet in the point L, which is the apex of the cone. 
Draw the axis L K, which produce in the direction 
of O, and at any convenient point upon the same con- 
struct a plan of the frustum at its top, a b. 

In connection with the profile 
of the cylinder draw a correspond- 
ing elevation of the cone, as shown 
by K 1 a 1 b 1 K 2 . Produce the sides 
K 1 «' and K s V until they intersect, 
thus obtaining the point L', the 
apex corresponding to L of the 
elevation. Draw the axis L 1 E, as 
shown, which produce in the direc- 
tion of N 1 , and upon it draw a 
second plan of the frustum at a b, 
as shown by M' O 1 N 1 . Divide the 
plans M N and M 1 O 1 N 1 into the 
same number of equal parts, com- 



duced merely to show how it may be done, should it 
be desired under similar circumstances in any other 
case. The development of the pattern in this case 
could be most easily accomplished by using L 1 as a 
center from which to strike arcs from the various 
points on the line a 1 K\ The same result is accom- 



s at corresponding points in 



mencmg 

each, as shown. With the T-square set parallel to 
the axis of the cones, and brought successively against 
the points in the plans, drop lines to the lines a b and 
a' b\ as shown. 

From L' draw lines through the points in a 1 b\ cut- 
ting the profile of the cylinder, as shown in K 1 E K\ 
and in like manner from the apex L draw lines indefi- 
nitely through the points in a b. Place the T-square 
parallel to the sides of the cylinder, and, bringing it 
against the points in the profile K 1 E K s just described, 
cut corresponding lines in the elevation, as shown at 
H K G. A line traced through these points of inter- 
section, as shown by H K G, will form the miter line 
between the two pieces as it appears in elevation. 

This miter line is not necessary in obtaining the 
pattern, but the method of obtaining it is here intro- 




Fig. 491. — The Pattern of a Frustum of a Cone Intersected at Its 
Lower End by a Cylinder, Their Axes Intersecting at Eight 
Angles. 

plished, however, by continuing the lines drawn from 
K 1 E K 2 until they meet the side a G of the cone pro- 
longed, as shown from G to Z. Thus a Z becomes in 
all respects the same as a 1 K 1 . 

From L as center, and with radius L a, describe 
the arc b~ 0?, upon which lay off a stretchout of the 
plan M O N of the frustum. Through each of the 
points in this stretchout draw lines indefinitely, radiat- 
ing from L, as shown. Number the points in the 



Pattern Problems. 



271 



stretchout a' b' corresponding to the numbers in the 
profile, commencing with the point occurring where it 
is desired to have the seam. Set the compasses to 
L Z as radius, and, with L as center, describe an arc 
cutting the corresponding lines drawn through the 
stretchout, as shown by 1, 5 and 1. In like manner 



reduce the radius to the second point in G Z, and de- 



scribe an arc 



cutting 2 



4, 4 and 2. Also bring the 



pencil to the third point and cut the lines correspond- 
ing to it in the same way. Then a line traced through 
the points thus obtained, as shown by H 1 K 3 G 1 , will 
be the pattern of the frustum. 



PROBLEM 141. 
The Pattern for a Conical Boss. 



The principles and conditions in this problem are 
exactly the same as those in the one immediately pre- 
ceding (that is, the frustum of a cone mitering against 




Fig. 492. — The Pattern for a Conical Boss. 

a cylinder, their axis being at right angles), but its pro- 
portions are so different that it is here introduced as 
showing that the same application of principles often 
produces results so widely differing in appearance as to 
be scarcely recognizable. 

Let A B C D of Fig. 492 represent the elevation 
of the boss that is required to fit against the cylindrical 



can, a portion of the plan of which is shown by the 
arc A B. The plan at the smaller end of the boss 
is represented by E F G H. Continue the lines A T> 
and B C until they intersect at K, 
which is the apex of the cone of 
which the boss is a frustum. An in- 
spection of the elevation will show that 
it is only necessary to describe one- 
fourth of the pattern, the remaining parts 
being duplicates. Divide one-quarter of 
the plan into any convenient number of 
parts, in the present instance four, as 
shown by the points in H E. Drop lines 
from these points to the base D C, as 
shown. Draw lines from K through the 
points in the base until they intersect 
the arc at A B. which represents the 
body of the can. These points can- 
be numbered to correspond with the 
points in the plan from which they are 
derived. At right angles to the line F K 
draw lines from the points on A B until 
they strike the line A K, where their 
true distances from K can be measured. 
With K as a center, and K D as radius, 
strike the arc L M N, equal in length to 
the circumference of plan. 

If the whole pattern of boss is to be 
described from measurements derived ' 
from elevation it will be necessary to 
reverse the order of the numbers for each 
quarter, as shown. From K draw lines 
extending outwardly through these 
points, as indicated by the small figures. With K as 
center, draw an arc from the point 1' until it inter- 
sects radial lines 1 drawn from K, as shown at 0, Z 
and R. In the same manner draw an arc from 2' to 
lines 2, &c, as shown. A line traced through these 
points will produce the desired patterns, as shown by 
LOSZVEN. 



272 



The New Metal Worker Pattern Booh. 



PROBLEM 142. 
Pattern for the Lip of a Sheet Metal Pitcher. 



Let A B C D of Fig. 493 represent the side ele- 
vation of a pitcher top having the same flare all around, 
and E F G H the plan at the base. By producing the 
lines A D and B C until they intersect in the point K, 
tlic apex of the cone of which the pitcher top is a sec- 
tion -will be obtained. Divide one-half of the plan into 
any convenient number of equal spaces, as shown by 
the points in G H E. From these points drop lines to 
the base D C, as indicated. Then draw radial lines 
from K, cutting the points in D C, and producing 
them until they intersect the curved line representing 
the top of the pitcher, otherwise an irregular cut 
through the cone, as shown by A B. For convenience 
in subsequent operations, number these points to cor- 
respond with the numbering of the points in the plan 
from which they are derived. 

Place the T-square at right angles to the axial line 
of the cone H K, and, bringing it against the several 
points in A B, cut the line D A, as shown in the 
diagram. By this means there will be obtained in the 
line K A the length of radii which will describe arcs 
corresponding to points in the top line of the lip A B. 
With K as center and K D as radius, describe the arc 
L M N, which in length make equal to the circumfer- 
ence of the plan by stepping off on it spaces equal to 
the spaces originally established in the plan, all as in- 
dicated by the small figures. From K through each 
of the points inLMN thus established draw radial 
lines, extending outwardly indefinitely, as shown. 
Then with K as center and K", K, 1 K", etc., as radii, 
strike arcs, which produce until they intersect radial 
lines of corresponding number just drawn, all as shown 
in the diagram. Then a line traced through the points 
thus obtained will be the required pattern, all as shown 
by L P R N M. 

The method above described is a strictly mathe- 
matical rule for obtaining such shapes when a design 
embodying the necessary curve at the top is at hand. 
As by the nature of the problem, this part of the pat- 
tern does not require to be fitted or joined to any 
other piece, it would be much easier to obtain, by the 
foregoing method, the principal points in the outside 
curve of the pattern and finish by drawing the re- 
mainder to suit the taste of the designer. In other 
words, after the arc L M N has been drawn and 
stepped off into spaces, draw radial lines from K 
through the points representing the highest and the 



lowest parts required in the top curve, as 0.5 and 12, 
upon which lines the required lengths can be set off. 
Then these points can be connected by any curve 
suitable for the purpose. 

The principle involved in the foregoing is exactly 
the same as that of a hip or sitz bath given in the fol- 




~i 

Fig. 49S.— Pattern for the Lip of a Sheet Metal Pitcher. 

lowing problem, the difference in the finished article 
being a matter of size and shape and not of principle. 
Thus the sides could be made less flaring by placing 
the point K much further away from the base line, the 
hight A D could be increased and the curve A B 
could be altered to one more desirable ; but the various 
steps necessary to perform the task would remain 
exactly the same. 



Pattern Problems. 

PROBLEM 143. 
Pattern for a Hip Bath of Regular Flare. 



273 



Let C B D E, in Fig. 494, be the elevation of the 
body of a hip bath having an equal amount of flare on 
all sides, the plan of which is a circle. In describing 



At any convenient distance above D draw J K 
parallel to G E to be used as a regular base upon which 
to measure the circumference of the cone. Parallel to 
J K draw F H, and from a center obtained on F II by 
prolonging the axis A X draw a half -plan of the f rustum ? 
as shown by F & H. Divide this half -plan into any 
convenient number of equal parts, and from the points 
thus obtained carry lines parallel to the axis until they 
cut the line J K, and from there extend them in the 
direction of the apex A, thus cutting the curved line 
B D. Place the ["-square parallel with J K, and bring- 
ing it against the several points in the curved line B D, 
cut the side E D, as shown. From A as center, with 




Firj. 494.— Pattern for a Hip Bath of Regular Flare. 



the pattern for the body it will be considered as a sec- 
tion of a right cone, the plane C E being at right 
angles to the axis and the base being represented by 
the curved line B D, as shown. The sides E D and 
C B can be extended until they meet at A. Then A 
will be the apex of a cone of which C B D E is a frus 
turn having an irregular base B D. 



A K as radius, describe the arc K 1 K", on which lay off 
a stretchout of either one-half or the whole of the plan, 
as may be desired. In this case a half is shown. From 
the extremities of this stretchout, as K 1 and K' 2 , draw 
lines to the center, as K' A and K7 A, and from the 
several points in the stretchout draw similar lines, as 
shown by 1, 2, etc. With one point of the dividers 



274 



The New Metal Worker Pattern Book. 



set at A bring the pencil point to the point D in the 
side A K, and with that radius describe an arc, which 
produce until it cuts the corresponding line 12 in the 
stretchout, as shown at D 1 . In like manner, bringing 
the pencil point up to the several points between D 
and E in the elevation, describe arcs cutting lines of 
corresponding numbers in the stretchout. Then a line 
traced through these intersections will form the upper 
line of the pattern. From A as center, with A E as 
radius, describe the arc C E 1 , cutting A K 1 and A 
K", as shown by C 1 and E 1 , forming the lower line of 
pattern. Then C B 1 D l E 1 will be half the pattern 
for the side of the hip bath. 



As a feature of design, the form produced in the 
pattern by a curved line B D drawn arbitrarily may 
not be entirely satisfactory. If, for instance, that part 
of the pattern lying between lines 9 and 12 should not 
appear as desired, it can be modified upon the pattern 
at will, as this edge of the pattern is not required to fit 
any other form. Such a modification is shown by the 
dotted lines a 1 K? of the pattern and a K of the eleva- 
tion. The foot of the tub is a simple frustum of a 
right cone, the pattern for which is obtained in the 
manner described in Problem 123. Different forms of 
bathtubs in which the flare is irregular will be found 
in Section 3 of this chapter. 



PROBLEM 144. 

The Envelope of a Frustum of a Right Cone Contained Between Planes Oblique to It? Axis. 

In Fig. 495, let F L M K represent the section of 
the cone the pattern for which is required. Produce 
the sides F L and K M until they meet in the point N, 
which is the apex of the cone of which F L M K is a 
frustum. Through 1ST draw N E. bisecting the angle 
LNM and constituting the axis of the cone, which pro- 
duce in the direction of D indefinitely. From K draw 
K II at right angles to the axis. At any convenient 
distance above the cone construct a plan or profile as it 
would appear when cut on the line K H, letting the 
center of the profile fall upon the axis produced, all as 
shown by ADC B. Divide the profile into any num- 
ber of equal parts, and from the points thus obtained 
draw lines parallel to the axis, cutting K H. From 
the apex N, through the points in K H, draw lines 
cutting the top L M and the base F K. Place the 
blade of the T-square at right angles to the axis of the 
cone, and, bringing it successively against the points 
in L M and F K, cut the side N F, as shown above L, 
and from H to F. From 1ST as center, with radius 
N II, strike the arc T S indefinitely, upon which lay off 
a stretchout from the plan, as shown, and through 
the points of which, from the center 1ST, draw lines in- 
definitely, as shown. "With the point of the conrpasses 
still at N, and the pencil brought successively against 
the points in the side from H to F, describe arcs, 
which produce until they eut corresponding lines drawn 
through the stretchout. Then a line traced through Fig ' 495 - The f T^* °J " *?£*" °f v */* C °"" Contained 

5 between Planes Oblique to its Axis. 

these points of intersection, as shown by T TJ S, will 

form the lower line of pattern. In like manner draw ing lines drawn through the stretchout. A line traced 

arcs by radii corresponding to the points in the side at through these points, as E P 0, will be the upper line 

L, which produce also until they intersect correspond- of the pattern sought. 




Pattern Problems. 



275 



PROBLEM 145. 



The Pattern of a Cone Intersected by a Cylinder at Its Upper End, Their Axes Crossing at Right Angles. 



In the plan, Fig. 496, let A B C D F represent a 
frustum of the cone B C G, B H C being the 
half profile of cone at its base and A D J the 
plan of the cylinder. In line with the cylinder in plan 
draw the elevation, as shown by R S T U. With the 
T-square placed parallel to the sides of the cylinder, 
carry a line from the point G in plan to any convenient 
point, as G' of elevation. At right angles to G G' 
draw G' Q indefinitely, and extend B C through Q 
G', cutting same at 0. With O as center, and E C 
of plan as radius, describe the semicircle L Q M, rep- 
resenting one-half of the profile of the cone at the 
larger end. Divide LQM into any convenient num- 
ber of equal parts, as indicated by the small figures. 
From the points thus obtained carry lines at right 
angles to L M, cutting that line as indicated. From 
the points thus obtained in L M carry lines to the apex 
G', as indicated by the dotted lines in the engraving. 
Divide BHC into the same number of equal parts as 
was LQM, numbering them to correspond with the 
elevation, as shown, and from the points thus obtained 
carry lines at right angles to B C, cutting that line as 
indicated. From the points in B C carry lines to the 
apex G, cutting the plan of cylinder as shown. With 
the blade of the T-square placed parallel with G G', 
and brought successively against the points thus es- 
tablished in the plan of the cylinder, cut lines of corre- 
sponding number drawn from the points in L M to 
the apex G', as indicated from K to 1ST, and extend 
these lines to the line M G'. A line traced through 
these points of intersection, as shown by K P 1ST, rep- 
resents the intersection of cone with cylinder in eleva- 
tion, as shown by A F D in plan. 

For the pattern proceed as follows : From G' as 
center, with G' M as radius, describe an arc, as indi- 
cated by I m, and, starting from I, step off the stretch- 
out of the half profile LQM, as indicated by the 
small figures. If the entire pattern is required in one 
piece extend the arc I o m, and from m set off a du- 
plicate of I m, numbering the points in inverse 
order. From the points thus obtained draw radial 
lines to G', as indicated. Then with G' as center, and 
with radii corresponding to the distance from G' to the 
points established in M G', describe arcs, producing 



them until they cut lines of corresponding number 
drawn from G'. A line traced through these points of 
intersection, as shown by k p n, will with I m give 




Fig. 496. — Pattern of a Cone Intersected by a Cylinder at its Upper End. 

the pattern of part of article shown in elevation by 
L M N P K. 

It will be easily seen that the pattern might have 



276 



The New Metal Worker Pattern Book. 



been obtained directly from the plan without the 
trouble of drawing the elevation, as in Problems 141, 
142 and 143. Should it be desirable, however, to cut 
an opening in the side of the cylinder to fit the frus- 
tum of the cone, the hights of all points in the perim- 



eter of such opening must be obtained from the line 
NP K of the elevation, while width of the opening 
upon lines corresponding to these points must be 
measured from F toward Dor A upon the circumfer- 
ence of the cylinder. 



PROBLEM 146. 

Pattern of a Tapering: Article with Equal Flare Throughout, which Corresponds to the Frustum of a 

Cone Whose Base Is an Approximate Ellipse Struck from Centers, the Upper Plane of 

the Frustum Being: Oblique to the Axis. 



In Fig. 497, let H F G A be the shape of the 
article as seen in side elevation. The plan is shown 
by I L N 0. In order to indicate the principle in- 
volved in J±e development of this shape, it will be 
necessary first to analyze the figure and ascertain the 
shape of the solid of which this frustum is a part. 
Since by the conditions of the problem the base is 
drawn from centers and the sides have equal flare, it 
follows that each arc used in the plan of the base is a 
part of the base of a complete cone whose diameter can 
be found by completing the circle and whose altitude 
can be found by continuing the slant of its sides till 
they meet at the apex, all of which can be seen by an 
inspection of the engraving. Thus those parts of the 
figure shown in plan by K U T M and E IT T P may 
be considered as segments cut from a right cone, the 
radius of whose base is either O K or L R, and 
whose apex E is to be ascertained by continuing the 
slant of the side L 2 C 1 till it meets a vertical line 
erected from O 1 of the plan, which is the center of the 
arc of the base, all as shown in the end view. Also 
those parts of the plan shown by K U R and M T P 
are segments of a right cone whose radius is U I or T 
]ST and whose altitude is found, as in the previous case, 
by continuing the slant of its side Gr A (which is par- 
allel to C 1 L 2 ) till it meets a vertical line erected from 
its center T, as shown in the side view. 

To complete the solid, then, of which F G A H is 
a frustum, it will only be necessary to take such parts 
of the complete cones just described as are included 
between the lines of the plan and place them together, 
each in its proper place upon the plan. The resulting 
figure would then have the appearance shown by H D 
CB A when seen from the side, and that of O a C L a 



when seen from the end. The lines of projection con- 
necting the various views together with the similarity 
of letters used will show the correspondence of parts. 
This figure is made use of in the second part of 
Chapter V, Principles of Pattern Cutting, to which 
the reader is referred for a further explanation of prin- 
ciples. 

Divide one-half of the plan into any convenient 
number of equal parts, as shown by the small figures, 
and from the points thus established carry lines verti- 
cally, cutting the base line H A, and thence carry them 
toward the apexes of the various cones from the bases 
of which they are derived. That is, from the points 
upon the base line H A derived from the arc K M 
draw lines toward the apex E, and from the points de- 
rived from the arc I K carry lines toward the apex D, 
and in like manner from the points derived from the 
arc M N carry lines in the direction of the apex B, all 
of which produce until they cut the top line F Gr of 
the article. From the points in F G thus established 
carry lines to the right, cutting the slant lines of the 
cones to which they correspond. Thus, from the 
points occurring between F and f draw lines cutting 
B A, being the slant of the small cone, as shown by 
the points immediately below W. In like manner, 
from the points between g and G- carry lines cutting 
the same line, as shown at G. The slant line of the 
large cone is shown only in end elevation, and there- 
fore the lines corresponding to the points between / 
and g must be earned across until they meet the line 
B 1 IA 

Commence the pattern by taking any convenient 
point, as E 1 , for center, and E 1 1/ as radius, and strike 
the arc L ! S indefinitely. Upon this arc. commencing 



Pattern Problems. 



277 



at any convenient point, as K 4 , set off that part of the 
stretchout of the plan corresponding to the base of the 
larger cone, as shown by the points 5 to 13 in the plan, 
and as indicated by corresponding points from K 4 to 
M 2 in the arc. From the points thus established draw- 
lines indefinitely in the direction of the center E', as 



as shown by/ 1 g\ Next take A B of the side eleva- 
tion as radius, and, setting one foot of the compasses in 
the point K' of the arc, establish the point D 1 in the 
line K 4 E 1 , and in like manner, from M 2 , with the same 
radius, establish the point B 2 in the line M* E', which 
will be the centers from which to describe those parts 
of the patterns derived from the smaller cones. From 
D 1 and B 1 as centers, with radius B A, strike arcs from 
K 4 and M 2 , respectively, as shown by K 4 F and M 2 N"', 
upon which set Off those parts of the stretchout cor- 
responding to the smaller cones, as shown by the arcs 
K I and M N of the plan. From the points thus estab- 
lished, being 5 to 1 and 13 to 17, inclusive, draw radial 
lines to the centers D 1 and B 2 , as shown. 




Fig. 497.— Pattern of the Frustum of a Cone, the Base of which is 
an Approximate Ellipse Struck from Centers, the Upper Plane of 
the Frustum being Oblique to the Axis. 

shown. From E 1 as center, with radii corresponding 
to the distance from E 1 to points 5 to 13 inclusive, 
established in the line B 1 L 5 already described, cut cor- 
responding radial lines just drawn, and through the 
points of intersection thus established draw a line, all 



For that part of the pattern shown from F 1 to /', 
set the dividers to radii, measuring from B, corre- 
sponding to the several points immediately below W of 
the side elevation, and from D 1 as center cut the cor- 
responding radial lines drawn from the arc. In like 
manner, for that part of the pattern shown from G- 1 to 
g\ set the dividers to radii measured from B, corre- 
sponding to the points in the line B A at G, with which, 
from B 2 as center, strike arcs cutting the correspond- 
ing measuring lines, as shown. Then F 1 Gr 1 W F will 
be one-half of the pattern sought — in other words, the 
part corresponding to I K L M N of the plan. The 
whole pattern may be completed by adding to it a du- 
plicate of itself. 



278 



The New Metal Worker Pattern Book. 



PROBLEM 147. 
The Envelope of a Right Cone, Cut by a Plane Parallel to Its Axis. 



Let B A F in Fig. 49S be a right cone, from which 
a section is to be cut, as shown by the line C D in the 
elevation. Let GLHK be the. plan of the cone in 
which the line of the cut is shown by D 1 D\ For 
the pattern proceed as follows : Divide that portion 
of the plan corresponding to the section to be cut off, 
as shown by D 1 G D 2 , into as many spaces as are nec- 
essary to give accuracy to the pattern, and divide the 
remainder of the plan into spaces convenient for laying 
off the stretchout. From A as center, with radius A 
B, describe an arc, as M N, which make equal to the 
stretchout of the plan G L II K, dividing it into the 
same spaces as employed in the plan, taking care that 
its middle portion, D 3 T>% is divided to correspond with 
D 1 D 5 of the plan. From the points in M N correspond- 
ing to that portion of the plan indicated by D' G D" — 
namely, 8 to 16 inclusive — draw lines to the center A, 
as shown. 

From points of the same number in the plan carry 
lines vertically, cutting the base of the cone, as shown 
from B to D, and thence continue them toward the apex 
A, cutting C D, as shown. From the points in C D 
carry lines at right angles to the axis A E cutting the 
side of the cone, as shown by the points between C 
and B. From A as center, with radii corresponding to 
the distances from A to the several points between C 
and B, cut lines drawn from points of corresponding 
number in the stretchout, to A, and through the 
points of intersection thus obtained trace a line, as 
shown by D 3 C 2 ~D*. Then the space indicated by D 3 
C 2 D* is the shape to be cut from the envelope MAN 
of the cone to produce the shape to fit against the line 
C D in the elevation. 

To obtain the pattern of a piece to fill the open- 
ing D 3 C 2 D 4 in the envelope, shown in profile by C D of 
the elevation, first draw radial lines from the center 
P of the plan to points 9 to 15 inclusive on the cir- 
cumference, cutting D 1 D\ Upon B F of the elevation 
extended, erect any perpendicular as D 6 C", as shown 




Fig. 



498.— The Envelope of a Right Cone Cut by a Plane 
Parallel to its Axis. 



at the left, which cross with horizontal lines projected 
indefinitely from the points on C D. The width of 
the piece upon each of these lines may be found by 
measuring upon D 1 D 2 of the plan the distances be- 
tween the intersections of radial lines of corresponding 
number, as between lines 11 and 13, 10 and 14, etc. 
A line traced through the points so obtained, as shown 
by D 6 C b D 7 , will give the desired pattern. 



PROBLEM 148. 
The Pattern for a Scale Scoop, Having: Both Ends Alike. 



In Fig. 499, let A B C D represent the side ele- 
vation of a scale scoop, of a style in quite general 



use, and E F II G a section of the same as it would 
appear cut upon the line B D, or, what is the same, 



Pattern Problems. 



279 



so far as concerns the development of the patterns, 
an end elevation of the scoop. The curved line ABC, 
representing the top of the article, may be drawn at 
will, being, in this case, a free-hand curve. For the 
patterns proceed as follows : From the center K, by 
which the profile of the section or end elevation is 




Fig. $99. — The Pattern for a Scale Scoop. 



drawn, draw a horizontal line, which produce until it 
meets the center line of the scoop in the point 0. 
Produce the line of the side ~D until it meets the line 
just drawn in the point X. Then X is the apex and 



X the axis of a cone, a portion of the envelope of 
which each half of the scoop may be supposed to be. 

Divide one-half of the profile, as shown in end 
elevation by E Gr, into any convenient number of 
spaces, and from the points thus obtained carry lines 
horizontally, cutting the line B D, as shown, and 
thence carry lines to the point X, 
cutting the top B C, as shown. 
With X D as radius, and from X 
as center, describe an arc, as shown 
by L 1ST, upon which lay off the 
stretchout of the scoop, as shown 
in end elevation. From the points 
in L X thus obtained draw lines 
to the center X, as shown. From 
the points in B C drop lines at right 
angles to X, cutting the side D C, 
as shown. With X as center, and 
radii corresponding to each of the 
several points between D and C, 
describe arcs, which produce until 
they cut radial lines of correspond- 
ing numbers drawn from points 
in the arc L N to the center X. Then a line traced 
through the points thus obtained, as shown by L M N, 
will be the profile of the pattern of one-half of the re- 
quired article. 



PROBLEM 149. 

The Patterns for a Scale Scoop, One End of Which Is Funnel Shaped. 



In Fig. 500 is shown a side view of a scale scoop 
by which it will be seen that the portion A B Gr H of 
the funnel-shaped end is a simple cylinder and, there- 
fore, need not be further noticed here. In Fig. 501 
are shown a side and an end elevation of the tapering 
portions. It will also be seen that the part ~D E F of 
the side view is similar in all respects to the article 
treated in the preceding problem, and the pattern 
shown in connection with the same is obtained by 
exactly the same method as that there described and 
need not, therefore, be repeated. 

An inspection of the side elevation will show that 
the part G B C D F is a section of a cone of which I 
is the apex, H F the base and H' C" F' the plan of 
the base, and that this cone is cut by the lines B Gr 
and C D. To obtain the pattern of this part, first 



divide F' C" and C" H' of end elevation into any con- 
venient number of parts, and from the points thus ob- 




Fio. 500. — Scale Scoop, One End of which is Funnel Shaped, 

tainea carry lines cutting the miter line H F, as 
shown. From the points in H F carry lines to the 



280 



Tlie New Metal Worker Pattern Book. 



apex I, cutting the curved line C D, as shown. From 
the points in C D drop perpendiculars cutting the sides 



the same as in Fig. 501. "With I of Fig. 502 as cen- 
ter, and I Gr and I F as radii, describe the arcs E, S 




Fig. 502— Pattern of Piece B C D F G of Fig. 500. 

G F, as shown. For convenience in describing the 
pattern a duplicate of the side view of this part is 
shown in Fig. 502, in which similar parts are lettered 



P Q. Upon P Q lay out twice 
stretchout of H' C" F' of Fig. 
if the pattern is desired in one 
piece. Thus the stretchout of H' C" 
is represented by P IT and Q "W, 
from the points in which draw lines 
to the center I. From I as center, 
and radii corresponding to the dis- 
tance from I to each of the various 
points in G F, describe arcs cutting 
lines of similar numbers. Trace lines 
through the points thus obtained, 
as shown by T TJ, V W. The ETUWVS is the 
pattern for part of scoop shown in side elevation by 
BCDFG. 



PROBLEM 150. 



The Pattern of a Conical Spire Mitering upon Four Gables. 



Let E I B in Fig. 503 be the elevation of a pin- 
nacle having four equal gables, down upon which a 
conical spire is required to be mitered, as shown. 



Produce the sides of the spire until they meet in the 
apex D. Also continue the side E F downward to any 
convenient point below the junction between the spire 



Pattern Problems. 



281 



and the gables, as shown by H, which point may be 
considered the base of a cone of which the spire is a 
part. Let YKLM be the plan of the gables. The 
diagonal lines V L and M K represent the angles or 
valleys between the gables, while R S and T U repre- 
sent the ridges of the gables over which the spire is to 
be fitted. Through the point H in the elevation draw 
a line to the center of the cone and at right angles to 
its axis, as shown by H 0. This will represent the 
half diameter or radius of the cone at its base. With 
radius C H, and from center A 2 of the plan, describe a 
circle, as shown, which will represent the plan of the 
cone at its base. 

At any convenient distance from the elevation, 
and to one side, project a diagonal section correspond- 
ing to the line M A 2 in the plan, as follows : From 
all the points in the side of the pinnacle draw horizon- 
tal lines indefinitely to the left, which will establish 
the bights of the corresponding points in the section. 
From any vertical line, as D' A', as a center line set off 
upon the horizontal lines the distances as measured 
upon the line M A 2 of the plan. Thus make B 1 A 1 




of the crossing of the two ridges of the gables, there- 
fore a line drawn from F 1 to B 1 will represent one of 
the valleys between the gables. Draw H 1 D 1 , the side 
of the cone. Its intersection with the line of the val- 
ley at G will then represent the hight of the lowest 
points of the spire between the gables, and a line pro- 




Fig. 504.— Pattern. 



M U L 

Fig. 503.— Plan, Elevation and Diagonal Section. 



The Pattern of a Conical Spire Mitering Upon Four Gables. 



equal to M A 2 and C B? equal to A 2 5, the radius of 
the cone at its base. The point F 1 represents the hight 



jected from this point back into the elevation, as shown, 
will locate those points in that view. 



282 



The New Metal Worker Pattern Book. 



To describe the pattern, first divide one-eighth of 
the plan of the cone, choosing the one which miters 
with the gable shown in the elevation, into any num- 
ber of equal spaces, as shown by the small figures. 
From these points carry lines vertically cutting the 
base of the cone H C, as shown, and thence toward the 
apex D, cutting the line B J of the gable, against which 
this part of the cone is to miter. As the true distance 
of any one of the points just obtained upon the line B 
J from the apex D can only be measured on a drawing 
when that point is shown in profile, proceed to drop 
these points horizontally to the profile line D H, where 
they are marked l 1 , 2 1 , etc., and where their distances 
from D can be measured accurately. Next draw any 
straight line, as D 2 H 2 of Fig. 504, upon which set off 
all the distances upon the line D II of the elevation, 
all as shown, each point being lettered or numbered 
the same as in the elevation. With D 2 of Fig. 504 as 
a center, from each of these points draw arcs indefi- 
nitely to the left, as shown. Upon the arc drawn from 
I!" set off spaces corresponding to those used in spac- 
ing the plan, beginning with IF, as shown by the small 
figures, and from each poiut draw a line toward the 



center D 2 cutting arcs of corresponding number drawn 
from the line F 2 H 2 . A line traced through the 
points of intersection (g to F 2 ) will give the shape of 
the bottom of the cone to fit against the side of 
one of the gables, or one-eighth of the complete 
pattern. 

By repeating the space 1 5 upon the arc drawn 
from H 2 seven times additional, as marked by the points 
1 and 5, the point V will be reached, from which a line 
drawn to D 2 will complete the envelope of the cone. 
From the points marked 1 and 5 draw lines toward D* 
intersecting the arcs of corresponding number. This 
will locate all of the highest and lowest points of the 
pattern, after which the miter cut from g to F 2 can be 
transferred by any convenient means, as shown from g 
to /, and so on, reversing it each time, as shown. In the 
case of a spire of very tall and slender proportions it 
will be sufficiently accurate for practical purposes to 
draw the lines g F 2 and g f straight. But the broader 
the cone becomes at its base the more curved will the 
line g F 2 become. With a radius equal to D E of Fig. 
503 describe the arc E 2 F, as shown, which will com- 
plete the pattern. 



PROBLEM 151. 



The Pattern of a Conical Spire Mitering Upon Eight Gables. 



In Fig. 505 is shown the elevation of a pinnacle 
having eight equal gables, upon which the conical 
spire E F P I is to be fitted. Produce the sides F E 
and P I until they meet in the point D, which is the 
apex of the spire. Let A IF S K M N'TU represent 
the plan of the pinnacle drawn in line just below the 
elevation. To ascertain the length of the cone forming 
the spire at its longest points, where it terminates in 
the valleys between the gables, it will be necessary to 
construct a section on the line A B representing one of 
the valleys in plan, which can be done as follows : 
From the points D, F and H in the elevation project 
lines horizontally to the left, which intersect with any 
vertical line, as D 1 B 1 , representing the center line of 
spire in the section. Upon the line drawn from H set 
off from B 1 a distance equal to A B of the plan and 



draw A 1 F 1 . From D' draw a line parallel to D F cut- 
ting A 1 F 1 in B 1 ; then D 1 R 1 Avill be the length or slant 
hight of the cone at its longest points, and a line from 
R 1 projected back into the elevation will locate the base 
of the cone in that view, as shown at R. 

From B as a center, with a radius equal to R 3 R 1 , 
describe a circle in the plan, which will represent the 
base or plan of the cone. Divide an eighth of this 
circle into any number of equal parts, as shown by 1, 
2, 3, 4 and 5, which spaces are to be used in measur- 
ing off the arc circumscribing the pattern. Draw any 
line, as D R in Fig. 506, upon which set off the sev- 
eral points in the line D R, as shown by the letters, 
and from D as center describe arcs indefinitely from 
each point, as shown. On the arc drawn from R step 
off spaces corresponding to one-eighth of the plan, as 



Pattern Problems. 



283 



shown by 1, 2, 3, 4 and 5. Draw the line D 5, as 
shown, cutting the arc from F, as indicated by/. By 




Fig. SOS. — Elevation and Plan of Conical Spire Mitering upon 
Eight Gables. 



inspection of the elevation it will be seen that the arc 
F represents the line of the top of the gables, and that 



the arc R represents the line of points in the base of 
the cone to fit down between the gables. Therefore 
from F to the middle point 3 draw F g, and from / 
draw/ g. Then ~D f g F will be one-eighth of the re- 
quired pattern. Set the dividers to 1 3 on the arc R 
and step off a sufficient number of additional spaces to 
complete the pattern, as shown by 3, 5, 3, 5, etc., to W. 
Draw W D. Also from the points 3 draw the lines 
f'g and gf, thus completing the pattern. 

In case the work is of large dimensions it will be 
advisable to miter the cone from F to g, g to/, etc., 
in the manner shown in the preceding problem, but in 




6 rvrr 



Fig. 506.— Pattern of Spire Shown in Fig. 505, 



case the work is small it will be sufficiently accurate to 
make the lines fg straight, as shown. 



z$i 



The New Metal Worker Pattern Book. 



PROBLEM 152. 



Patterns for a Two-Piece Elbow in a Tapering: Pipe. 



In the solution of this problem two conditions may 
arise ; in the first, the two pieces of the elbow have the 
same flare or taper, while in the second case one of the 



both would present to view two perfect ellipses of ex- 
actly the same proportions and dimensions ; and, there- 
fore, that if the two parts be placed together again, 




Fig. 507.— A Two-Ptece Elboiv in a Tapering Pipe. 



pieces may have more flare than the other. It has 
been shown in the chapter on geometrical problems that 
an oblique section through the opposite sides of a cone 
is a perfect ellipse. Keeping this in mind, it is evident 
that if the cone shown by A B C in Fig. 507 were 
made of some solid material and cut obliquely by the 
plane D E and the severed parts placed side by side, 



turning the upper piece half way around, as shown by 
D E A 1 , the edges of the two pieces from D to E would 
exactly coincide. 

Taking advantage of this fact, then, it only becomes 
necessary to ascertain the angle of the line D E, neces- 
sary to produce the required angle between the two 
pieces of an elbow, both of which have equal flare. 



Pattern Problems. 



285 



Therefore, at any convenient point upon the axis A H, 
as I, draw I J at the angle which the axis of the upper 
piece is required to make with that of the lower, then 
bisect the angle J I H, as shown by the line I K. 
Draw D E parallel to I K at the required hight of the 
lower piece, which will be the miter line sought. 

Before completing the elevation of the elbow it 
will be necessary to notice a peculiarity of the oblique 
section of a cone — viz., that although the line A H 
bisects the cone and its base it does not bisect the 
oblique line D E, as by measurement the center of D E 
is found to be at x. Therefore, through the point b, 
which is as far to the right of x as point a is to the left 
of it, draw any line, as b A 1 , parallel to I J and make b 
A 1 equal in length to a A, and draw A' D and A 1 E. 
Next draw G 1 F 1 at right angles to A 1 b, representing 
the upper end of the elbow. Make D F equal to E F', 
and E G equal to D G 1 . Then B F G C will be the 
elevation of a frustum of a cone, which, when cut in 
two upon the line D E, will, when the upper section is 
turned half-way around upon the lower part, form the 
elbow BDG'FEC. 

At any convenient distance below the base of cone 
B C draw half the plan, as shown by L H M, which 
divide into any convenient number of equal spaces. 
From the points of division erect lines vertically, cut- 
ting the base of the cone B C, and thence carry them 
toward the point A, cutting the miter line J) E. Plac- 
ing the T-square parallel to the base line B C bring it 
successively against the points in D E, cutting the side 
of the cone, as shown below D. 

From A as center, with radii A B and A F, draw 
arcs, as shown. Upon the arc drawn from B, begin- 
ning at any convenient point, as N, step off a stretch- 
out of L H M, as shown by the small figures. From 
each of the points thus obtained draw measuring lines 



toward the point A, and from the last point one 
cutting the arc drawn from F at Q. Placing one point 
of the compasses at the point A, bring the pencil point 
in turn to each of the points in the side of the cone 
below D and cut measuring lines of corresponding 
number. Then a line traced through the points of in- 
tersection, as shown from S to R, will be the miter cut 
between the two parts of the pattern of the frustum 
1ST P Q necessary to form the patterns of the required 
elbow. 

As but half the plan of the cone was used in ob- 
taining a stretchout, the drawing shows but halves ol 
the patterns. In duplicating the halves to form the 
complete patterns the upper piece can be doubled upon 
the line Q S and the lower part upon the line R N, 
thus bringing the joints on the short sides. 

If, according to the second condition stated at the 
beginning of this problem, the upper section of this 
elbow is required to have more or less flare than the 
lower section, thereby placing the apex A 1 nearer to or 
farther away from the line D E, a different course will 
have to be pursued in obtaining the pattern. If, for 
instance, the hight of the cone A H be reduced, the 
base B G remaining the same, the proportions — that is, 
the comparative length and width — of the ellipse derived 
from the cut D E would be different from those derived 
from the same cut were the proportions of the cone to 
remain unchanged. Therefore, since the shape of the 
lower piece at the line D E is a fixed factor, if the 
circle at G 1 F 1 be shifted up or down the axis, or, re- 
maining where it is, its diameter be changed, the piece 
D G 1 F' E becomes an irregular tapering article, in 
which case its pattern can most easily be obtained by 
triangulation. Patterns for pieces embodying those 
conditions can be found in Section 3 of this chapter, 
to which the reader is referred. 



PROBLEM 153. 



Patterns for a Three-Piece Elbow in a Tapering Pipe. 



In Fig. 508 is shown a three-piece elbow occur- 
ring in taper pipe, in which the flare is uniform 
throughout the three sections. In solving this problem 
the simplest method will be to construct the elevation 
of the elbow and an elevation of an entire cone, from 
which several sections may be cut to form the re- 



quired elbow, at one and the same time. Therefore 
in the elevation of the cone E F G let L 1 M 1 be drawn 
at a distance from E F equal to the total length of the 
three pieces measured upon their center lines, and 
also let its length be equal to the diameter of the 
elbow at its smaller end ; then through E and L 1 and 



286 



Hie New Metal Worker Pattern Book. 



through F and M 1 draw the sides of the cone, inter- 
secting in G. 

At any convenient point, asB, draw the line B A 
at the angle which the axis of the middle piece is 
required to make with that of the lower (in this case 
45 degrees), and bisect the angle A B G, as shown, by 
the line B D. Parallel with B I) draw P B at any re- 



upper piece is required to make with that of the 
middle piece (in this case also 45 degrees, or hori- 
zontal), and bisect the angle S U I, as shown, by U T. 
Then the miter line 1ST can be drawn parallel with 
U T at any required distance from I, upon which 
locate the point i, making 1ST i equal to j. From i 
draw the axis of the upper piece of the elbow paral- 



K . — -""" 


L— -"Vx 




\ / 




Fig. 508. — A Three-Piece Elbow in a Tapering Pipe. 



quired hight, upon which locate the point I, making P I 
equal to B J. (The reason for this is explained in the 
previous problem.) From the point I draw the axis 
of the second section of the elbow parallel with A B, 
making it (I H) equal to J G, and draw P H and R H. 
From any convenient point upon this axis, as 17, 
draw US at the required angle which the axis of the 



lei with 17 S, making i K equal to j H. Next locate 
the line N upon the original cone, making N 1 P 
equal to R and 1 R equal to P N. Now make 
N L equal to N 1 L and M equal to M' O 1 and draw 
ML. 

It may be remarked here that on account of the 
shifting of the positions of the axes of the several 



Pattern Problems. 



287 



pieces upon the miter lines by turning them, as shown 
by I J and i j, it will be impossible to ascertain with 
extreme accuracy the lengths of the various pieces 
upon their axes until the elevation EPNLMORF 
is drawn, and therefore to obtain the position which 
the line M L will occupy. 

This method of solving the problem is given upon 
the supposition that its simplicity will compensate for 
this slight inaccuracy, as usually differences of length 
can be made up in the parts with which the elbow may 
be connected. If the lines M L and E F are to be 
assumed at the outset as fixed factors between which 
a tapering elbow is to be constructed, it will be some- 
what difficult to ascertain the exact dimensions of a 
cone, E F G, which can be cut and its parts turned so 
as to constitute the required elbow. Hence, while 
two of the pieces (say the two lower ones) can easily 
be cut from an entire cone assumed at the outset, the 
third piece will have to be drawn arbitrarily to fit be- 
tween the last miter line N and the small end M L, 
and will very likely be of different flare from that of 
the other two pieces. This will necessitate the last 
section being cut by the method of triangulation, 
problems in which are demonstrated in Section 3 of 
this chapter, to which the reader is referred. 



Having, as explained above, obtained the lines 
of cut through the cone, the patterns may be described 
as follows : Draw the plan V W Y, its center X fall- 
ing upon the axis of the cone produced, which divide 
in the usual manner into any convenient number of 
equal parts. Through the points thus obtained erect, 
perpendiculars to the base E F, and thence carry lines 
toward the apex G, cutting the miter lines P E and 
N' 0'. With the T-square at right angles to the axis 
G C, and brought successively against the points in 
N' 1 and P K, cut the side G F of the cone, as shown 
by the points above O 1 and below K. From G as center,, 
with radius G F, describe the arc E 1 F', upon which lay 
off the stretchout of the plan V W Y, as shown by the 
small figures 1, 2, 3, etc., and from these points draw 
measuring lines to the center G. From G as center 
describe arcs corresponding to the distance from G to 
the several points established in G F, which produce 
until they intersect lines of corresponding numbers 
drawn from the center G to the arc E 1 F'. Through 
these points of intersection trace lines, as shown by 
s N s and P' R 1 . From G as center, with radius G M 1 , 
describe the arc 1/ M 1 . Then L 2 M" N 2 O 2 is the half 
pattern of the upper section, O 2 N 3 E, 1 P 1 that of the- 
middle section, and P' E, 1 F 1 E 1 that of the lower section. 



PROBLEM 154. 



The Patterns for a Regular Tapering; Elbow in Five Pieces. 




b c 

Fin. 509.— Diagram of Angles for a Five-Piece Elbow. 



In this problem, as in the two immediately preced- 
ing, the various pieces necessary to form the elbow 
may be cut from one cone, whose dimensions must be- 
determined from the dimensions of required elbow. 
The first essential will be to determine the angle of the- 
cutting lines, which may be done the same as if the 
elbow were of the same diameter throughout. 

Such an elbow of five pieces would consist of three 
whole pieces and two halves ; therefore, if it is to be a 
right angle elbow, divide any right angle, as A' B C in 
Fig. 509, into four equal parts, as shown by the points 
1, 2, 3. Bisect the part A' B 3 by the line A B and. 
transfer the portion A' B A to the opposite side of the 
figure, as shown by C B C. 

This gives the right angle ABC divided into the- 
same number of pieces and half-pieces as would be em- 
ployed in constructing an ordinary five-piece elbow. 



288 



The New Metal Worker Pattern .Book. 




Fig. BIO. — A Five-Piece Elbow in a Tapering Pipe. 



Pattern Problems. 



289 



The division lines in this diagram are of the correct 
angle for the miter lines in the elbow pattern, and there- 
fore can be used npon the diagram of the cone, out of 
which are to be obtained the pieces to compose the re- 
quired elbow. 

It is assumed that the amount of rise and projec- 
tion are not specified, therefore after having got the 
line of the angle or miter it becomes a matter of judg- 
ment upon the part of the pattern cutter what length 




Fig. 511. — Elevation of Five-Piece Tapering Elbow. 



the 



shall be given to each of the pieces composing 
elbow. 

In Fig. 510, let A B represent the diameter of the 
large end of the elbow. From the middle point in the 
line A B, as 0, erect a perpendicular line, as indicated 
by ON, producing it indefinitely. On the line N, 
proceeding upon judgment, as already mentioned, set 
off G X to represent the length of the first section of 
the elbow measured upon its center line. With X thus 
determined, draw through it the line D E, giving it the 
same ansrle with A B as exists between B C of Fis<-. 
509 and the horizontal B C. This, in all probability, 
can most readily be done by extending B A indefinitely 



beyond A and letting E D intersect with B A extended, 
producing at their intersection an angle equivalent to 
CBC'of Fig. 509. From the point X set off the dis- 
tance X Y, also established by judgment, thus deter- 
mining the position across the cone of the miter line of 
the next section. Through Y draw G F at the same 
angle as D E, already drawn, but inclined in the oppo- 
site direction. In like manner locate the two other 
miter lines shown in the diagram, finally obtaining the 
point Z. From Z set off the width toward N of the 
last section of the pattern, and through the point N 
thus obtained draw the line M at right angles to C N, 
making it in length equal to the diameter of the small 
end of the elbow and placing its central point at JN . 
Through the points A M and B of the figure thus 
constructed draw lines, which produce indefinitely until 
they intersect the axis in the point P. Then P will be 
the apex of the required cone. 

Construct a plan of the base of the cone or large 
end of the elbow below and in line with the diagram, 
as shown in the drawing, which divide into any con- 
venient number of spaces, as indicated by the small 
figures, and from the points thus obtained carry lines 
vertically, cutting the base of the cone A B. From 
A B continue them toward the apex of the cone, cut- 
ting the several miter lines drawn. "With the apex P 
of the cone for center, and with P B as radius, describe 
the arc T U, upon which set off a stretchout of one- 
half the plan, all as indicated by the small figures. 
From the points thus established in T U carry lines to 
the center P. With the T-square placed at right angles 
to the axis N C of the cone, and brought against the 
points of intersection in the several miter lines made 
by the lines drawn from points in the base of the cone 
to the apex, cut the side B of the cone, as shown 
Then from P as center, with radii corresponding to the 
distance from P to the several points on B, as men- 
tioned, strike arcs cutting the lines of corresponding 
numbers in the pattern diagram, as shown. Then lines 
traced through the points thus obtained, as indicated by 
I) 1 E', F 1 G 1 , etc., will cut the pattern OWU T of the 
frustum in such a manner that the sections will consti- 
tute the half patterns of the pieces necessary to form 
the required elbow. In Fig. 511 is shown an elevation 
of the elbow resulting from the preceding operation. 



290 



Tlie New Metal Worker Pattern Book. 



PROBLEM 155. 

The Frustum of a Cone Intersecting; a Cylinder of Greater Diameter than Itself at Other than 

Right Angles. 



In Fig. 512, E Gr H F represents an elevation of 
the cylinder, and M N L K an elevation of the frustum 
of a cone intersecting it. FZQ represents the profile 
or plan of the cylinder, to which it will be necessary to 
add a correctly drawn plan of the frustum before the 
miter line in elevation cau be obtained. At any con- 



obtained drop points parallel with the side G H of the 
cylinder and continue them indefinitely, cutting the 
line F 1 O 1 , which is drawn through the center of the 
plan of the cylinder at right angles to the elevation, all 
as shown in the engraving. Make Y 1 W equal to Y W 
of the first section constructed. In like manner measure 




Fig. 512.— The Frustum of a Cone Intersecting a Cylinder of Greater Diameter than Itself at Other than Right Angles. 



venient point on the axial line T of the cone construct 
the profile V Y X W, which represents a section 
through the cone on the line M 1ST. Divide the section 
TYX AY into any convenient number of equal spaces 
in the usual manner, as shown by the small figures 1, 
2, 3, 4, etc. From each of the points thus established 
drop lines parallel with the axis of the cone cutting 
the line M N. From the intersections in M N thus 



distances from the center line V X of the first section 
to the points 2, 3, 4, etc., and set off corresponding 
spaces in the plan view, measuring from M 1 N 1 , upon 
lines of corresponding numbers dropped from the in- 
tersections in M N", already described. Then a line 
traced through these points will represent a view of the 
upper end of the frustum as it would appear when 
looked at from a point directly above it. Produce the 



Pattern Problems. 



291 



sides of the frustum K M and L N" until they meet in 
the point 0. From drop a line parallel to the side 
G H of the cylinder, cutting the line F' 0' in the point 
O', thus establishing the position of the apex of the 
cone in the plan. From the point O 1 thus established 
draw lines through the several points in the section 
M 1 Y 1 N 1 W, which produce until they intersect the 
plan of the cylinder in points between Z and Q, as 
shown in the engraving. From 0, the apex of the cone 
in the elevation, draw lines through the several points 
in M 1ST already determined, which produce until they 
cross G H, the side of the cylinder, and continue them 
inward indefinitely. Intersect these lines by lines drawn 
vertically from the points of corresponding number be- 
tween Z and Q of the plan just determined. Then a 
line traced through these intersections, as indicated by 
K T L, will represent the miter between the frustum 
and cylinder, as seen in elevation. 

To lay off the pattern proceed as follows : From 
O as center, with X as radius, describe the arc P R, 
on which set off a stretchout of the section YV¥X 
in the usual manner. From 0, through the several 
points in P R thus obtained, draw radial lines indefi- 
nitely. From the several points in the miter line K T L 



draw lines at right angles to the axis T of the cone, 
producing them until they cut the side N L. From O as 
center, with radii corresponding to the distance from to 
the several points in N L just obtained, describe arcs, 
which produce until they intersect radial lines of corre- 
sponding number drawn through the stretchout P R. 
Then a line traced through these points of intersection, as 
indicated by S 1/ U, will be the lower line of the pat- 
tern sought, and PSL'UE will be the complete pattern. 
The pattern for the cylinder and the opening in 
the same to fit the intersection of the cone is really a 
problem in parallel forms, with which problems (Section 
1) it should properly be classed. F 1 Z Q is the pro- 
file of the cylinder, and L T K is the miter line. The 
stretchout B D is drawn at right angles to E F, the 
direction of the mold or cylinder. The points between 
Z~ and Q ! of the stretchout are duplicates of those be- 
tween Z and Q of the plan. Place the T-square at 
right angles to the cylinder, and, bringing it successively 
against the points in the miter line K T L, cut lines of 
corresponding numbers. A line traced through the 
points of intersection thus formed, as shown by Z 1 K 1 
Q 1 L', will be the shape of the required opening in the 
cylinder. 



PROBLEM 156. 

The Patterns of the Frustum of a Cone Joining: a Cylinder of Greater Diameter than Itself at Other 
than Right Angles, the Axis of the Frustum Passing - to One Side of That of the Cylinder. 



Let EJHG in Fig. 513 be the elevation of a 
cylinder, which is to be intersected by a cone or frus- 
tum, D AJC, at the angle F D A in elevation, and 
which is to be set to one side of the center, all as 
shown by S P L M R of the plan. Opposite the end 
of the frustum, in both elevation and plan, draw a sec- 
tion of it, as shown by TUTW in the elevation and 
T 1 U' V W in the plan. Divide both of these sec- 
tions into the same .number of equal parts, commenc- 
ing at corresponding points in each, and number them 
as shown by the small figures in the diagram. From 
the points in T II V ¥ carry lines parallel to the axis 
of the cone, cutting the line A J, and thence drop 
them vertically across the plan. From the points in 
the section T 1 II 1 V 1 W draw lines parallel to the axis 
of the cone, as seen in plan, intersecting the lines of cor- 
responding number dropped from A J just described. 
Through these points of intersection trace a line, as 
shown by L M. Then L M will show the end of the 



frustum A J as it appears in plan. From X, the 
apex of the cone in elevation, drop a line vertically, 
cutting the axis of the cone in plan as shown at X 1 . 
From X draw lines through the points in A J and 
extend them through the side of the cylinder indefi- 
nitely. From X 1 through the points in L M draw 
lines cutting the plan of the cylinder, as shown from 
P to R, and from these points carry lines vertically, 
intersecting those of corresponding number in the 
elevation drawn from the apex X. Tnen a line traced 
through these points, as shown by D K N, will be 
the miter line in elevation. 

For the pattern of the frustum, from X as center, 
with radius X A, describe the arc A 1 J 1 , upon which 
lay off a stretchout of the section T IT V W, through 
the points in which, from X, draw radial lines indefi- 
nitely. From the points in D K G 1ST carry lines at 
right angles to the axis of the cone, cutting the side 
A D extended, as shown from D to B. From X as 



292 



Tlte New Metal Worker Pattern Book. 



center, with radii corresponding to the distance from 
X to the various points in the line D B, describe arcs 
cutting radial lines of corresponding number in the 
pattern. Through the points of intersection in the 



tern of the frustum D A J C, inhering with the cylin- 
der at the angle described. 

The method of obtaining the pattern of the cyl- 
inder is analogous to that described in the preceding 




Fig. 513.— The Patterns of the Frustum of a Cone Joining a Cylinder of Greater Diameter than Itself at Other than Right Angles, 
the Axis of the Frustum Passing to One Side of that of the Cylinder. 



pattern thus obtained trace a line, as shown by 
C K 1 N 1 D'. Then D 1 N 1 K' C A' J 1 will be the pat- 



problem, and is clearly shown at the left in the 
drawing. 



PROBLEM 157. 

The Patterns of a Cone Intersected by a Cylinder of Less Diameter than Itself, Their Axes Crossing; at 

Right Angles. 



In Fig. 514, let B G B D F A C be the elevation 
of the required article. Draw the plan in line with the 
elevation, making like points correspond in the two 



views, as shown by M S T U P K Let D M E be a 
half section of the cylinder in the elevation and D 1 M 1 
E 1 a corresponding section in the plan. 



Pattern Problems. 



293 



Divide these sections into any convenient number 
of equal parts, commencing at the same point in each, 
as shown by the small figures, and draw the center line 
of the cylinder in plan D 1 R. From each of the points in 
the section shown in elevation carry lines parallel to D 

H 




Fig. 514. — A Cone Intersected by a Cylinder of Less Diameter than 
Itself at Right Angles to Its Axis. 

F cutting the side of the cone, and extend them some 
distance into the figure for further use. From the 
several points of intersection with the side of the cone, 
as shown by a, b, c, d and e, drop lines parallel to the axis 



of the cone cutting the line D 1 R of the plan, giving the 
points a l ,b l ,c l ,d x and e', and through each of these points, 
from R as center, describe an arc, as indicated in the en- 
graving. From the points in the profile D 1 M' E 1 of 
the plan draw lines parallel to the sides of the cylinder, 
producing them until they meet the arcs drawn through 
corresponding points, giving the points indicated by 
l 1 , 2', 3 1 , 4 1 and 5'. From these points carry lines 
vertically to the elevation, producing them until they 
meet the lines drawn from points of corresponding 
numbers in the profile of the cylinder in the elevation, 
giving the points l a , 2% 3', 4" and 5". A line traced 
through these points, as shown from Gr to F, will be 
the miter line in elevation formed by the junction of 
the cvlinder and the cone. 




Fig. 515.— Half Pattern of the Cone Shown in Fig. 514. 

To obtain the envelope of the cone with the open- 
ing to fit the intersecting cylinder proceed as follows : 
From any convenient point, as A 1 , Fig. 515, draw A 1 
B 1 , in length equal to A B of the elevation. Set off 
points e", d', c", b" and a~ in it, corresponding to e, d, c, 
b and a of A B, Fig. 514. From A' as center, with 
radius A 1 B 1 , describe the arc B' V, upon which lay off 
the stretchout of the plan of the cone, as indicated by 
the small figures outside of the pattern. (But one-half 
of the envelope of the cone is shown in the engraving.) 
From the same center A 1 describe arcs from the points 
e% d 2 , c 2 , V and a". From the center R of the plan draw 
lines to the circumference through the points 2 1 , 3 1 , 4', 
etc. , giving the points in the circumference marked 2 s , 
3 3 , 4 3 , etc. Set off by measurement corresponding 
points in the arc B' V, as showu by 3 4 , 2 4 , 4 4 , 5 4 , etc. 
From these points draw lines to the center A 1 , inter- 
secting the arcs of corresponding number drawn from 



294 



Tlie New Metal Worker Pattern Book. 



a?, b\ (?, etc. A line traced through these points of 
intersection, as shown by F 1 0' (x P', will be the shape 
of the opening to be cut in the side of the cone to 
fit the mitered end of the cylinder. 



The pattern for the cylindrical part is shown above 
the elevation, and is obtained in accordance with the 
principles demonstrated in the first section of this chap- 
ter, which need not be here repeated. 



PROBLEM 158. 

The Patterns of a Cone Intersected by a Cylinder of Less Diameter than Itself at Right Angles to its 
Base, the Axis of the Cylinder Being to one Side of that of the Cone. 



N 4 



N 3 




PATTERN OF CYLINDER 

In Fig. 516, let B A C represent the elevation of 
the cone, DE GrH the elevation of the cylinder, which 
joins the cone at right angles to the base BC. J K 
L M N P Q is the plan of the articles, which is to 
be drawn in line and under the elevation, making like 
points correspond in the two views, as shown. Draw 
a section of the cylinder in line with the elevation, as 
shown by E F G E. Divide the section of the cylinder 
into any convenient number of equal parts, as shown 
by the small figures. From the apex A drop a line 
through the plan, as shown by A M. Through the 
center of the section of the pipe, as shown in plan, 
draw a straight line to the center of plan of cone, as 
shown by J P. This line will also be at right angles 
to K M. From each of the points in the section of the 
pipe in elevation drop lines parallel to the sides of the 
pipe cutting the side of the cone, extending them to 
the line J P in plan, as shown by 1ST a b c, etc. Through 
each of these points, from P as center, describe circles, 




PLAN 



Fig. 516.— Plan and Elevation of Cone Intersected by a Cylinder at 
Right Angles to its Base. 



Pattern Problems. 



295 



as shown, cutting the sides of the plan of cylinder. 
From each of the points of intersection with the side 
of the cone (A B) draw lines parallel with the base, and 
extend them inward. If it is desired to show the miter 
line in elevation formed by the junction of pipe and 
cone, from the points d ef in the plan of cylinder carry 
lines vertically to the elevation, producing them until 
they meet the horizontal lines having similar letters 
drawn through the side of the cone A B, giving the 
points g h j. A line traced through these points, as 
shown by D g hj H, will be the miter line. 



H f e d D of elevation, as shown by H c b a D. From 
the center A of pattern describe arcs cutting the points 
HckD. 

It is only necessary now to make each of these arcs 
equal in length to the one to which it corresponds in the 
plan by any method most convenient. Thus make a d 
and a d l equal to a d of the plan, b e and b e 1 equal to 
b e of the plan and c/and cf equal to cfoi the plan. 
A line traced through these points, as shown by H Q 
D 0, will be the shape of the opening. Another method 
of making the measurements of the arcs is shown by 





Fig. 617. — Half Pattern of Cone Shown in Fig. 516. 



Fig. 51S.— Perspective View of Cone and 
Cylinder Sliown in Fig. 516. 



The half pattern of the cone, with the opening to 
fit the cylinder, is shown in Fig. 517, to describe which 
proceed as follows : From any convenient point, as A 
in Fig. 517, with A B of Fig. 516 as radius, strike an 
arc indefinitely, as shown. From B of pattern set off 
each way the stretchout of J M and J K of plan and 
connect K and M with A. Then K A M B is the half 
pattern of the cone, or as much as shown on plan by 
K J M. To obtain the shape of opening to be cut in 
cone to correspond with the shape of pipe, on A B, the 
center line of pattern, set off points corresponding to 



the radial dotted lines of the plan and pattern in the man- 
ner explained in the problem immediately preceding. 

The pattern for the cylinder is obtained in the 
manner usual with all parallel forms, its only pecul- 
iarity in this case being that its stretchout is taken 
from the irregular spaces upon the profile NOPQ oi 
the plan, which are transferred to the line P' P", as 
shown. 

A pictorial representation of the finished article is 
shown in Fig. 518, upon which some of the lines oi 
measurement shown in Fi°\ 516 have been traced. 



296 



Tlie New Metal Worker Pattern Book. 



PROBLEM 159. 

Patterns of a Cylinder Joining: a Cone of Greater Diameter than itself at Right Angles to the Side 

of the Cone. 



Let B A K in Fig. 519 be the elevation of a right 
cone, perpendicular to the side of which a cylinder, L 
S T M, is to be joined. The first operation -will be to 
describe the miter line as it would appear in elevation. 
Draw the section TJ V of the cylinder, which divide 
into auy convenient number of equal parts, as indicated 
by the small figures, and from these points drop lines 
parallel to L S, cutting the side A K of the cone in the 
points H, F and D, producing them until they cut the 
axis A X in the points G, E and C. In order to ascer- 
tain at what point each of these lines will cut the en- 
velope of the cone it will be necessary to construct 
sections of the cone as it would appear if cut on the 
lines G H, E F and C D. Draw a second elevation of 
the cone, as shown by B 1 A 1 K', representing the cone 
turned quarter way round ; the first may be regarded 
as a side elevation and this as an end elevation. Draw a 
plan under the side elevation of the cone, as shown by 
NEPO, which divide into any convenient number of 
equal parts, and in like manner draw a corresponding 
plan or half plan under the end elevation, as shown by 
K 1 P 1 O 1 . Divide this second plan into the same spaces, 
numbering them to correspond with the other plan. 
From the points 1 to 5 in plan NEPO carry lines 
vertically to the base B K and thence toward the apex 
A, cutting the lines C D, E F and G H. In like man- 
ner, from the same points (1 to 5 inclusive) in the plan 
R' P 1 O 1 carry vertical lines to the base B 1 K 1 and 
thence toward the apex A 1 . Place the T-square at right 
angles to the axes of the two cones, and, bringing it 
against the points of intersection of the lines from X to 
K with C D, cut corresponding lines in the second ele- 
vation, and through the points of intersection thus es- 
tablished trace a line, as shown by C 3 C 4 . Produce the 
axis X 1 A 1 to any convenient distance, upon which set 
off C D 1 , in length equal to C D, in which set off the 
points corresponding to the points in C D, and through 
these points draw lines at right angles to C D 1 . Place 
the T-square parallel to the axis X 1 A 1 , and, bringing 
it against the several points in C 3 C 4 , cut the lines of 
corresponding number drawn through C 1 D', as shown, 
and through the intersections thus established trace a 
line, as shown. Then C D 1 is a section of the cone as 
it would appear if cut on the line C D. 

In like manner carry lines from the points upon 



E F across to the end elevation, intersecting them with 
lines of corresponding number, as shown from E 3 to E 4 , 
and thence carry them parallel to the axis, cutting lines 
drawn through E 1 F 1 , which with its points has been 




Fig. 519.— A Cylinder Joining a Cone of Greater Diameter than Itself 
at Right Angles to the Side of the Cone. 

made equal to E F. The resulting profile E 1 F' is a 
section of the cone as it would appear if cut on the line 
E F. Also use the points in G H in like manner, es- 



Pattern Problems. 



297 



tablishing the profile G' H 1 , which represents a section 
of the cone as it would appear if cut on the line G H. 
(Some of the lines indicating the operation in connec- 
tion with sections E 1 F 1 and G 1 H 1 are omitted in the 
engraving to avoid confusion.) 

Having thus obtained sections of the cone corre- 
sponding to the several lines C D, E F, G H, it will 
next be necessary to project a plan of the cone, with 
its intersecting cylinder, at right angles to L S, or as 
viewed in the direction of A K, which plan shall in- 
clude all of these sections. To do this extend the line 




Fig. 520. — Envelope of Cone Shown in Fig. 519. 



A K to a convenient distance above the elevation, and 
project lines from all other important points parallel to 
the same, as shown. At right angles to A K draw any 
line, as C~ V, as a center line of the new plan. As the 
points D 1 , F' and H 1 of the oblique sections of the cone 
are all in the line A K, transfer these sections to the 
new plan, so placing them that their center lines shall 
coincide with the center line of the new plan, and the 
points D', F 1 and H 1 shall be at the intersection of A 
K with the center line of the plan, all as shown. Op- 
posite the end of the cylinder draw a section, as indi- 
cated by U 1 V, which divide into the same number of 
equal parts as used in the divisions of U V, commenc- 
ing the division at corresponding points in each. As 
both halves of the cylinder and of the cone, when di- 
vided by a vertical plane passing through the axis of 
each, are the same, only one-half of the section of the 
cylinder has been numbered. From the points in U 1 
V drop lines parallel to C 2 V, each line cutting its 



corresponding section, as shown from x toy, and then 
carry them parallel to A K back to the elevation, cut- 
ting lines of corresponding number in that view. That 
is, from the intersection of the line drawn from point 
4 in U 1 V with the profile C 2 L 3 cut the line C D, 
which in the elevation corresponds to the point 4 in 
the profile U V, and from the intersection of e line 
drawn from 3 with E 2 L 3 cut the line E F, and so on, 
all as indicated by the dotted lines. Then a line traced 
through these points of intersection, as shown by L M r 
will be the miter line in elevation, from which the pat- 
terns may readily be obtained. 

The intersections in the plan above give all that is 
necessary to obtain the pattern of the cylinder, which 
can be done as follows : Lay off a stretchout of the 
profile U 1 V opposite the end S 2 T 2 , through the point- 
in which draw the usual measuring lines. Place 
the T-square at right angles to the same, and, bringing 
it against the points in the miter line L M (or the points 
of intersection in x y in the plan from which L M was 
obtained), cut the corresponding measuring lines. Then 
a line traced through these points, as shown from L' to 
M 1 , will be the shape of the pattern of the cylinder to 
fit against the cone. 

For the pattern of the cone proceed as follows -. 
From each of the points in the miter line L M carry 
lines horizontally across, cutting the side A B of 
the cone, by means of which their distance from the 
apex A may be accurately measured; also through 
these points draw lines from the apex A cutting the 
base B K, continuing them vertically into the plan N 
B P O,. as shown. It may be noted that the line from 
point 4 on L M falls at point 2 in the plan of the cone \ 
likewise that the line from 3 on L M falls at 3 in the 
plan of the cone, while the line from 2 falls upon the 
plan of the cone at a point marked a. From any 
convenient point, as A 2 of Fig. 520, with a radius equal 
to A B, describe the arc B 2 K 2 B 3 , which in length 
make equal to the circumference of the plan of the 
cone, setting off in the same all the points of the plan, 
as indicated by the figures and letters, and from these 
points draw lines toward the center A 2 , all as shown. 
From A 2 as center, with radii corresponding to the dis- 
tances A L, A 2, A 3, A 4 and A M of the elevation, 
strike arcs intersecting corresponding lines just drawn. 
Then a line traced through the intersections thus ob- 
tained will be the shape of the opening to be cut in the 
envelope of the cone to fit the end of the cylinder. 



^98 The Neiv Metal Worker Pattern Book. 

PROBLEM 160. 

The Patterns of a Cylinder Joining the Frustum of a Cone in which the Axis of the Cylinder is 
Neither at Right Angles to the Axis Nor to the Side of the Cone. 



The principles involved in the solution of this 
problem are exactly the same as those of the problem 
immediately preceding, to which the reader is referred 
for a more full explanation of the operation. The de- 



section of the cylinder, which divide into any con- 
venient number of equal parts, as indicated by the 
small figures, and from these points carry lines parallel 
with N X cutting the side B L of the cone in the 




Fig. 521. — A Cylinder Joining the Frustum of a Cone at an Oblique Angle. 



tails or conditions differ only in the angle at which the 
cylinder joins the side of the cone. 

In Fig. 521, let C B L be the elevation of a right 
cone of which C c I L is a frustum, and let M T U N 
represent the cylinder which is to join the frustum., 
making the angle TJ E" L greater than a right angle. 
The first operation will be to determine the shape of 
miter line M N of side elevation. Draw V W X, the 



points A, G and E, producing them until they cut the 
axis B Y in the points H, F and D. Draw a plan un- 
der the side elevation of cone, as shown by S Q P, 
which divide into any convenient number of equal 
parts. From points 1 to 4 in S Q P carry lines verti- 
cally to the base C L, and thence toward the apex B, 
cutting the lines D E, F Gr and H A 

Draw a second elevation of the cone, as shown by 



Pattern Problems. 



2J9 



C B 2 L 1 , which represents the cone as turned quarter 
way round. Draw a corresponding plan under the end 
elevation, as shown by S 1 Q 1 P 1 O 1 . Divide this plan 
into the same number of equal parts, commencing to 
number them at the same point as in the other plan — 
that is, at the point Q. From the points 1 to 4 in- 
clusive in S 1 Q 1 P' of plan carry vertical lines to the 
base C 1 L', and thence to the apex B 2 . 

The next step is to construct sections of the cone 
as "it would appear if cut upon the planes represented 
by the lines Ii A, F G and D E. For this purpose 
place the J-square at right angles to the axes of the 
two cones, and, bringing it against the points of in- 
tersection of the lines from the base G L with D E, 
cut corresponding lines in the second elevation, and 




Fig. 522. — Envelope of Cone Shown in Fig. 521. 

through the points of intersection thus established 
trace a line, as shown by N 1 E 3 X 2 . 

Continue the axis Y 1 B 2 as may be convenient, 
upon which set off spaces equal to those between the 
points in D E, and through these points draw lines at 
right angles to D 1 E 1 . Place the T-square parallel to 
the axis Y 1 B 2 , and, bringing it against the several 
points in N 1 E 3 N 2 , cut the lines drawn through D 1 E 1 , 
as shown, and through these points of intersection 
trace a line, as shown by W E 1 N 3 . Then W E 1 N 3 is 
a section of the cone as it would appear if cut on the 
line D E. Sections corresponding to F Gr and H A 
can be obtained in a similar manner. 

Having obtained sections of the cone correspond- 
ing to the several lines D E, F G and H A, it will next 



be necessary to project a plan at right angles to the 
axis of the cylinder, in which each of these sections 
shall find its place. Therefore, from all the points of 
the cylinder and of its intersections with the sides and 
axis of the cone project lines at right angles to N X 
indefinitely, through which at any convenient point 
draw a line, as D 2 X', parallel to X X. Upon this line, 
as a center of the plan about to be constructed, place 
the oblique sections just obtained so that each may be 
in line with the line in the elevation which it repre- 
sents, and their center lines shall all coincide with 
D 2 X 1 , all as shown. Make T 2 IF equal to T U and 
complete the plan of the cylinder, opposite the end 
of wliich draw a profile, as indicated by V W X', 
commencing the divisions at the point V. From the 
several points in the profile V 1 AY X 1 drop lines paral- 
lel with the center line D 2 X' against the several 
profiles d E 2 d\f G Q /' and h A 2 A 1 , and thence drop 
the points back on the elevation, cutting correspond- 
ing lines in it. Thus, from the intersection of the 
line drawn from point W (3) with G 2 / 1 of section cut 
the line F G, which in the elevation corresponds to 
the point 3 in the profile Y W X. From the intersec- 
tion of a line drawn from point 2 in V 1 W with A 2 h 1 
of section cut the line H A, and so on, as indicated 
by the dotted lines. A line traced through these 
points of intersection, as shown by the curved line 
M N, will be the miter line in elevation, from which 
the patterns can be obtained as follows : 

For the pattern of cylinder shown in elevation by 
M T U X, on T U extended lay off a stretchout of 
profile V W X, through the points in which draw the 
usual measuring lines. Place the J-square parallel 
with T U, and, bringing it against the points in the 
miter line M X, cut measuring lines of corresponding 
number. Trace a line through the points thus ob- 
tained, as shown from m to m'. Then m t t' w! is the 
pattern of the cylinder to fit against the cone, as shown 
in elevation by MTUN. 

To obtain the pattern of the frustum carry lines 
from each of the points in the miter line M N horizon- 
tally across the elevation, cutting the side of the frus- 
tum c C. as shown by a\ b' and d 1 ; also through the 
same points draw lines from the apex B, cutting the 
base line C L, and thence drop them on the plan, as 
shown by 1, 2, a and b. From any convenient point, 
as B 2 in Fig. 522, as a center, with radii equal to B c 
and B C, describe arcs, as shown by Q 1 and c 1 V. 
Make OQO' equal in length to the plan of cone S 
Q P and upon it set off each way from the point Q 



500 



Tlie New Metal Worker Pattern Book. 



spaces equal to those upon the plan between Q and S. 
From these points draw lines indefinitely toward the 
center B 2 . With B 2 as center describe arcs whose 
radii correspond to B M, B a\ B b\ B d 1 andB 1ST, cutting 
lines of corresponding number or letter. Then a line 



traced through the intersections thus obtained will be 
the shape of opening to cut in the envelope of frustum 
where it joins the cylinder, and lines drawn from O 
and O 1 toward B 2 till they cut the arc c 1 1 1 in the points 
c 1 and V will complete the pattern of the frustum. 



PROBLEM 161. 

The Patterns of Two Cones of Unequal Diameter Intersecting: at Rig-lit Angles to their Axes. 



Let U T V in Fig. 523 be the elevation of a cone, 
at right angles to the axis of which another cone or 
frustum of a cone, F G P, is to miter. Lex L K 1ST M 
be a section of the frustum on the line F G. Let U 2 
W V s be a half plan of the larger cone at the base. 
The first step in describing the patterns is to obtain the 
miter line in the elevation, as shown by the curved line 
from to P. With this obtained the development of 
the pattern is a comparatively simple operation. 

To obtain the miter line P proceed as follows : 
Divide the profile L K N M into any convenient num- 
ber of equal parts, as shown by the small figures. In- 
asmuch as the divisions of this profile are used in the 
construction of the sections — or, in other words, since 
sections through the cone must be constructed to cor- 
respond to certain lines through this profile — it is de- 
sirable that each half be divided into the same number 
of equal parts, as shown in the diagrams. Thus 2 and 
2, 3 and 3, 4 and 4 of the opposite sides correspond, 
and sections, shown in the upper part of the diagram, 
are taken upon the planes which they represent. From 
the points in the profile LKN M draw lines parallel 
to B E cutting the end F G of the frustum. Produce 
the sides F and P G until they meet in E, which is 
the apex of the cone. Through the points in F G draw 
lines from E, producing them until they cut the axis 
of the cone, as shown at A, A 1 , A 2 . 

Next construct sections of the cone as it would 
appear if cut through upon the lines A C, A 1 B, A 2 D. 
Divide the plan U 2 W V 2 into any convenient number 
of parts. From the points thus established carry lines 
vertically to the base line TJ V, and thence carry them 
toward the apex T, cutting the lines A C, A 1 B, A 2 D, 
all as shown. Through each of the several points of 
intersection in these lines draw horizontal lines from 
the axis of the cone to the side, all as shown. At right 
angles to the lines A C, A 1 B, A 2 D project lines to 
any convenient point at which to construct the re- 
quired sections. Upon the lines drawn from the points 



A, A 1 , A 2 locate at convenience the points A 3 , A 1 , 
A 6 . Inasmuch as A 1 B is at right angles to the axis 
of the cone, the section corresponding to it will be a 
semicircle whose radius will be equal to A 1 B. There- 
fore, from A 3 as center, with radius A 1 B, describe the 
semicircle S B 1 P. For the section corresponding to 
A 2 D lay off from A 6 the distances A" S 2 and A 6 K'' 3 in 
a line drawn at right angles to A 2 D of the elevation, 
each in length equal to the horizontal line drawn 
through the point A 2 from the axis to the side of the 
cone. At right angles to S 2 R 2 draw A 6 D 1 , in length 
equal to A 2 D of the elevation. Set. off in it points 5, 
4 and 2, corresponding to similar points in A'" D of the 
elevation. Through these points 5, 4 and 2, at right 
angles to A 6 D 1 , draw lines indefinitely. From A 6 as 
center, with radius equal to the length of horizontal 
line passed through point 5 in A 2 D of the elevation, 
describe an arc cutting line 5 of the section. From the 
same center, with a radius equal to the length of the 
horizontal line drawn through point 4 in the line A 2 D 
of the elevation, strike an arc cutting the line 4, etc. 
Then a line traced through these points, as shown by 
S 2 D 1 R 2 , will be the section of the cone as it would 
appear if cut on the line A 2 D of the elevation. In like 
manner obtain the section S 1 C B. : , corresponding to 
A C of the elevation. 

These sections may, if preferred, be obtained in 
the manner described in connection with Problems 159 
and 160. 

As these sections are obtained solely for the pur- 
pose of determining at what point in their perimeters — 
that is, at what distance from points C, B 1 and D' — they 
will be intersected by the lines representing the points 
2, 3 and 4 of the profile L K M N", it is not necessary 
that the complete half sections should be developed. 
In the engraving, the small intersecting cone has so 
little flare that the lines A C and A 2 D cross the large 
cone so nearly at right angles to its axis that sections 
2 2 and 4 4 could be constructed with sufficient accu- 



Pattern Problems. 



30} 




Pig. 525.— The Patterns of Two Unequal Cones Intersecting at Right Angles to their Axes. 



302 



77ie New Metal Worker Pattern Book. 



racy for practical purposes, as in the case of section 3 3, 
by small arcs of circles with radii respectively equal to 
A C and A 2 D, and of only sufficient length to include 
the points c c and d d. 

Prolong A 5 D 1 , as shown by E 3 , making A s E 3 in 
length equal to A' E of the elevation. In like manner 
make A 4 E 2 and A 3 E' equal to A E and A" E of the 
elevation respectively. At right angles to these lines 
in the sections set off F" G% F 9 G 3 , F 4 G 4 , in position 
corresponding to F G of the elevation. Make the 
length oi Y~ G" equal to the length across the section 
of the frustum marked 2 2. In like manner make F 3 
G 3 equal to 3 3, and F' G 4 equal to 4 4 of the section. 
From E', E* and E 3 respectively, through these points 
in the several sections, draw lines cutting the oblique 
sections just obtained. From the several points of in- 
tersection between the lines drawn from E 1 , E 2 , E 3 and 
the sections of the cone, as shown by d d, c c and b b, 
carry lines back to the elevation, intersecting the lines 
A C, A 1 B, A 3 D. Then a line traced through these 
several intersections, as shown from to P, will be 
the miter line in elevation. 

Having thus obtained the miter line, proceed to 
describe the patterns, as follows : For the envelope of 
the small cone, from E as center, with radius E G, de- 
scribe the arc F 1 G 1 , upon which set off the stretchout 
of the section LKMK Through the points in this 
arc, from E, draw radial lines indefinitely. From E as 
center, with radii corresponding to the several points 
in the miter line P, but obtained from the oblique 



sections above, cut corresponding radial lines. Thus 
with the radius E 3 d cut lines 4 and 4, with the radius 
E 2 c cut lines 2 and 2 and with radius E 1 b cut lines 3 
and 3. 

Then a line traced through these points of intersec- 
tion, as shown by P 1 0' P 3 , will be the shape of the 
pattern of the frustum to fit against the larger cone. 

For the pattern of the larger cone, from T as 
center, with radius T U, describe the arc V'U', in length 
equal to the circumference of the entire plan of the 
cone. From the points in the miter line P carry 
lines parallel to the base of the cone cutting its side 
T IT, as shown between O 3 and P 3 , also through the 
points in P draw lines from the apex cutting the base 
and thence carry them vertically to the plan. 

These points can be numbered upon the side of 
the cone to correspond with the plan, but entirely in- 
dependent of the system of numbers employed upon 
the smaller cone. Upon the arc V U 1 set off points 
corresponding to the points just obtained in the plan 
from the miter line, from which draw lines toward the 
center T. With one foot of the compasses set at the 
point T, bring the pencil point successively to the 
points between O 3 and P 3 and cut radial lines of corre- 
sponding number in the pattern. Then a line traced 
through these intersections, as shown by X Y Z Y 1 , 
will be the shape of the opening to be cut in the en- 
velope of the larger cone, over which the smaller cone 
will fit, and T U 1 V will be the envelope of the entire 
cone. 



PROBLEM 162. 



The Patterns of the Frustums of Two Cones of Unequal Diameters Intersecting: Obliquely. 



In Fig. 524, let M K P O be the side elevation of 
the larger frustum and F' G 1 S E the side elevation 
of the smaller, the two joining upon a line between 
the points E and S, which line must be obtained be- 
fore the patterns can be developed. Produce the sides 
S G 1 and E F 1 until they meet in the point E. At any 
convenient place on the line of the axis of the smaller 
frustum draw the profile H F K G, corresponding to 
the end F 1 G 1 . Divide this profile into any convenient 
number of equal parts, as shown by the small figures 
1, 2, 3, etc., and from these divisions, parallel to the 
axis of che cone, drop points on to F 1 G 1 . From the 
apex E, through these points in F 1 G 1 , carry lines 



cutting the side F P of the larger frustum, and pro- 
ducing them until they meet the center line, or the 
base P, all as shown by B A, C A 1 and D A 3 . 

The next step is to construct sections of the larger 
frustum as it would appear if cut on each of these 
lines, from which to obtain points of intersection with 
the lines of the smaller frustum for determining the 
miter line from E to S in the elevation. Draw the 
plan of the base of the larger frustum, as shown bv 
T U V W, and divide one-half of it in the usual man- 
ner. From these points carry lines vertically to tn& 
base P of the frustum. Produce the sides M and 
P N" until they meet in the point L. From the points 



Pattern Problems. 



303 



in the base line obtained from the plan carry lines 
toward the apex L, cutting the section lines A B, 



project lines at right angles cutting A 3 B 1 , as shown, 
in the points 4, 3 and 2. In like manner make A* C 




Fig. 524- — Patterns of the Frustums of Two Unequal Cones Intersecting Obliquely. 



A' C and A 3 D, as shown. Parallel to A B and of the 
same length, at any convenient point outside of the 
elevation, draw A 3 B', and from the points in A B 



equal and parallel to A C, and from the points m A C 
project lines at right angles to it, cutting it as shown, 
giving the points 4, 3 and 2. Also make A s D 1 equal 



•304 



The New Metal Worker Pattern Book. 



to the section line A" D of the elevation, and cut it by 
lines from the points in A 2 D, obtaining the points 
3 and 2, as shown. In order to complete these sev- 
eral sections, the width of the frustum through each 
of the points indicated is to be set off on correspond- 
ing lines drawn through A 3 B 1 , A 4 C and A 5 D'. To 
obtain the width through these points first draw an 
end elevation of the larger frustum, as shown by M' N 1 
P 1 0'. Produce the sides, obtaining the apex L'. 
Draw a plan and divide it into the same number of 
spaces as that shown in T U V W, and commence 
numbering at a corresponding point, all as indicated 
by V U' T 1 W. From the points in the plan carry 
lines vertically to the base O 1 P 1 , and thence toward 
the apex L'. Place the blade of the J-square at right 
angles to the axis of the cone, and, bringing it succes- 
sively against the points in the section line A B in the 
side elevation, draw lines cutting the axis of the end 
elevation, and cutting the lines corresponding in num- 
ber to the several points in A B, all as shown by a a, 
h h and c t. Make the length of the lines drawn 
through A 3 B equal to the corresponding lines thus 
obtained, as shown by a 1 a', b 1 b 1 , c 1 c 1 and d' d', and 
through these extremities trace a line, as shown by 
d 1 W d\ which will be the section through the cone 
when cut on the line A B. In like manner complete 
the sections i' C I 1 and /' D' /'. 

As remarked in the previous problem, it is only 
necessary that these sections should be developed far 
enough from the points B 1 , C and D 1 to receive the 
lines representing the sections of the smaller frustum. 
Produce A 3 B', making B 1 E 1 equal to B E of the ele- 
vation, and B 1 X 3 equal to B X 1 of the elevation. In 
like manner make C E 2 equal to C E, and C X' equal 
to G X. Make D 1 E 3 equal to D E, and D' X 5 equal 
to D X' 2 . Through X, at right angles to B 1 E", draw 
a line in length equal to the line 2 2 drawn across the 
profile FKG IT, with which this section corresponds, 
as shown by 2' 2'. Through X 4 draw a line equal to 
II K, as shown by H 1 K', and through X & draw 4' 4', 
in length equal to the line 4 4 drawn through the 
profile F K G H. From E', through the extremities 
of 2' 2', draw lines cutting the section. In like man- 
ner draw lines from E 2 through the points H 1 K 1 , and 
from E 3 through the points 4' 4'. From the points at 
which these lines meet the sections, as a 2 a~ in the first, 
o o in the second and m m in the third, carry lines 
back at right angles to and cutting the corresponding 
section lines in the elevation. A line traced through 
the points thus obtained, as shown by E S, is the 



miter line in elevation formed by the junction of the 

two frustums. 

Having thus obtained the miter line in elevation, 
proceed to develop the patterns as follows : From the. 
points in E S, at right angles to A 1 E, which is the 
axis of the smaller cone, draw lines cutting the side 
E S, as shown by the small figures 1, 2, 3, 4 and 5. 
These points are to be used in laying off the pattern 
of the smaller frustum. From E as center, with 
radius E G 1 , describe the arc F 2 G 2 , upon which step 
off the stretchout of the profile F K H G, numbering 




Fig. 525.— The Envelope of the Larger Frustum Shown in Fig. 52$. 

the points in the usual manner. Through the points, 
from the center E, draw radial lines indefinitely. 
From the same center, E, with radius E 1 (of the 
points in E S), cut the radial line numbered 1, and in 
like manner, with radii E2, E3, etc. , cut the correspond- 
ing numbers of the radial lines. A line, E 1 S 1 , traced 
through the several points of intersection thus formed 
will be the larger end of the pattern for the small frus- 
tum, thus completing the shape of that piece, all as 
shown by E 1 S' G 2 F\ 



Pattern Problems. 



305 



To avoid confusion of lines, the manner of ob- 
taining the envelope of the large frustum is shown in 
Fig. 525, which is a duplicate of the side elevation 
and plan shown in Fig. 524, the miter line E 1 S' and 
the points in it being the same. Similar letters refer 
to corresponding parts in the several figures. From 
Jj as center, with radius L" 0", describe an arc, as 
shown by Y Z, and from the same center, with radius 
L 2 M 2 , describe a second arc, as shown by y z. Draw 
Y y, and upon Y Z lay off the stretchout of the plan 
V s W T 2 , all as shown. Draw Z z. Then ZzyY 
will be the envelope of the large frustum. Through 
the points in the miter line E, 1 S 1 draw lines from the 
apex of the cone to the base, and from the base con- 
tinue them at right angles to it until they meet the 



circumference of the plan. Mark corresponding points 
in the stretchout Y Z, and insert any points which do 
not correspond with points already fixed therein. 
From each of the points thus designated draw a line 
across the envelope already described to the apex, as 
shown by 3 L", x 1/ and 1 IA Also, from the points 
in the miter line R 1 S' draw lines at right angles to the 
axis of the frustum cutting the side L" O 2 , as shown. 
From L 2 as center describe arcs corresponding to each, 
of these points and cutting the radial lines drawn 
across the envelope of the cone. A line traced through 
the points of intersection between arcs and lines of 
the same number, as shown by h Br A 1 S 2 , will be the 
shape of the opening to fit the base of the smaller 
frustum. 



;;ik, 



The New Metal Worker Pattern Booh. 



SECTION 3. 

Irregoflar Forms. 

(TRIANGULATION.) 



The class of subjects treated in this section will 
include all irregular forms which can be constructed 
from sheet metal by simple bending or forming, but 
whose patterns cannot be developed by the regular 
methods employed in the two previous sections of this 
chapter. These problems divide themselves naturally, 
in regard to the arrangement of the triangles used in 
the development of the patterns, into two classes, viz. : 
First, those in which the vertices of the triangles used 
in constructing the envelopes all terminate at a com- 
mon point or apex, and, second, those in which the 
relative position of the base and the vertex is reversed 
in each succeeding triangle, or, in other words, those 
in which the vertices of alternate triangles point in 
opposite directions. 

In the introduction to Section 2 (page 240), at- 
tention is called to the difference between a scalene or 
oblique cone and a right cone with an oblique base. 
The scalene cone may be called the type or representa- 
tive of a large number of forms belonging to the first 



class above mentioned, since many rounded surfaces 
entering into the construction of various irregular flar- 
ing articles are portions of the envelope of a scalene 
cone. The principles involved in this particular class 
of forms are explained in that part of Chapter V refer- 
ring to Fig. 271, page 94. 

Inasmuch as triangulation is resorted to in all cases 
where regular methods are not applicable, it is not sur- 
prising that the forms here treated, especially those 
included in the second class above referred to, are more 
varied in character than those of any other class to be 
met with in pattern cutting. An explanation of methods 
and principles governing these will be found in the 
third subdivision of Chapter V, beginning on page 86. 
The last few problems of this class are devoted to the 
development of the horizontal surfaces of arches in cir- 
cular walls. 

The arrangement of problems in this section will be 
in accordance with the above classification although no 
headings will be introduced to distinguish the classes. 



PROBLEM 163. 

The Envelope of a Scalene Cone. 



In Figs. 526 and 527 are shown perspective rep- 
resentations of scalene or oblique cones. In Fig. 526 
the inclination of the axis to the base is so great that 
a vertical line dropped from its apex would fall out- 
side the base, while in Fig. 527 a perpendicular from 
its apex would fall at a point between the center and 
the perimeter of its base. 

Supposing the circumference of the base in either 
case to be divided into a number of equal spaces, it is 
plain to be seen that lines drawn upon the surface 
of the cone from the points of division to the apex 
would be straight lines of unequal lengths, and that 
such lines would divide the surface of the cone into 
triangles whose vertices are at the apex of the cone 
and whose bases would be the divisions upon the base 
of the cone. It will be seen further that with the 
means at hand of determining the lengths of these 
lines forming the sides of the triangles, the pattern 



cutter possesses all that is necessary in developing 
their envelopes or patterns. 




Fig. 526. Fig. 527. 

Scalene Cones of Different Inclinations. 

In Fig. 528, D AHis an elevation of the cone 
shown in Fig. 526 and DGH is a half plan of the 



Pattern Problems. 



307 



same, drawn for convenience, so that D H is at once 
the base line of the elevation and the center line of 
the plan. Fig. 529 shows an elevation and plan of the 
cone shown in Fi°\ 527, drawn in the same manner. 
The principle involved in the development of the pat- 
terns of the two oblique cones is exactly the same and, 
as will be seen, letters referring to similar parts in the 
two drawings are the same; therefore the following 
demonstration will apply equally well in either case. 
From the apex A drop a perpendicular to the base 




Fig. 528.— Pattern of Cone Shown in Fig. 526. 

line, locating the point 1ST. Divide the base DGH 
into any convenient number of equal spaces, as shown 
by the small figures, and from the points thus ob- 
tained draw lines to the point N. These lines will 
form the bases of a series of right angled triangles of 
which A 1ST is the perpendicular hight, and whose hy- 
pothenuses when drawn will give the correct length of 
lines extending from the points of division in the base 
of the cone to the apex. The most convenient method 
of constructing these right angled triangles is to trans- 
fer the distances from N to the various points upon 



the circumference of the base to the line NDasa base 
line, measuring each time from the point 1ST, by which 
method the line A 1ST becomes the common perpen- 
dicular of all the triangles. Therefore from IS" as 
center, with the distances N 1, N 2, etc., as radii, de- 
scribe arcs as shown in the engraving, cutting the base 
line N D. Lines from each of these points to the 
apex, as A 1, A 2, etc., will be the required hypoth- 
enuses. 

The simplest method of developing the pattern is 
to first describe a number of arcs whose radii are re- 
spectively equal to the various hypothenuses just 
obtained; therefore place one foot of the compasses at 
A, and, bringing the pencil point successively to the 




Fig. 529.— Pattern of Cone Shown in Fig. 527. 

points 1, 2, 3, etc., upon the line N D, describe arcs 
indefinitely. From any point upon the arc drawn from 
point 1, as n, draw a line to A as one side of the j^at- 
tern. Next take between the feet of the dividers a 
space equal to the spaces upon the circumference of 
the plan, and placing one foot of the dividers at the 
point n, swing the other foot around till it cuts the 
arc drawn from point 2 ; then A n 2 will be the first 
triangle forming part of the envelope or pattern. 
With the same space between the points of the dividers, 
and 2 of the pattern as center, swing the dividers 
around again, cutting the arc drawn from point 3. 
Repeat this operation from 3 as center, or, in other 
words, continue to step from one arc to the next, until 
all the arcs have been reached, as at g, which in this 



308 



The New Metal Worker Pattern Book. 



case will constitute one-half the pattern ; after which, 
if desirable, the operation of stepping from arc to arc 
may be continued, as shown, finally reaching the point 



d. Draw d A and trace a line through the points ob- 
tained upon the arcs, as shown by n g d, which will 
complete the pattern. 



PROBLEM 164. 

The Envelope of an Elliptical Cone. 



In Fig. 530 is shown an elevation and plan of a 
cone whose base is an elliptical figure. So far as the 
solution of this problem is concerned the plan may be 
a perfect ellipse or an approximate ellipse drawn by 
any convenient method. Fig. 531 shows a perspective 
view of the cone in question, upon which lines have 
been drawn from points assumed in its base to the 



Fig. 530— Plan and Elevation of Cone with Elliptical Base. 

apex. From an inspection of this view it will appear 
that these lines, as in the case of the scalene cone, are 
of unequal length, and therefore that the pattern of its 
envelope may be developed by a method analogous to 
that adopted in the preceding problem. 

Since the cone consists of four symmetrical quar- 
ters, it will be necessary to obtain the envelope of only 
one quarter, from which the remainder of the pattern 
can be obtained by reduplication. Therefore draw a 
half side elevation, as shown by Y X G of Fig. 532, 
immediately below which draw a quarter plan, X C E, 



so that the line X C shall be common to both views, as 
shown. Divide E G into any convenient, number of 





Fig. 531. — Perspective View of C>>ne Shown in Fig. 530 with Lines 
Drawn upon its Surface Used in Developing Pattern. 

equal parts, as indicated by 1 the small figures. Lines 
drawn from the points in E C to X will give the base 
lines of a set of triangles, whose altitudes are equal to 




QUARTER PLAN 
Fig. 532.— Pattern of One-Half the Cone Shown in Fig. 530. 

the hight of the article X Y, and whose hypothenuses 
will give the true distances from the apex to the points 



Pattern Problems, 



30& 



assumed in tile base line. A convenient method for 
drawing these triangles is as follows : With X as cen- 
ter strike arcs from the points in E G, cutting X C, as 
shown by the small figures. Lines drawn from the 
points thus obtained to Y, as shown, will give the hy- 
pothenuses of the triangles. With Y as center, and the 
distance from Y to the several points in X as radii, 
strike arcs indefinitely. From Y to any point upon the 
arc draw any line, as Y E, which will form the edge 
of pattern corresponding to X E of plan. With the 



dividers set to the space used in stepping off E C of 
plan, and starting from E of the pattern, space off the 
stretchout of the plan, stepping from one arc to the 
next, as shown. From the point 6, or C 1 , draw a line 
to Y. Through the points thus obtained trace a line. 
Then Y G E is the pattern for that part of the article 
shown on plan by E X 0. This quarter can he du- 
plicated by any means most convenient so as to obtain 
the pattern for one-half or for the whole envelope in 
one piece, as desired. 



PROBLEM 165. 
Pattern for a Raised Boiler Cover With Rounded Corners. 



The shape of the cover considered in this problem 
may perhaps be more accurately described as that of 



ELEVATION 




Fig. 533. — Plan and Elevation of Cover. 

an oblong pyramid with rounded corners, as shown by 
the plan and elevation in Fig. 533, an inspection of 



which will also show that the rounded corners are por- 
tions of a scalene cone, while the four pyramidal sides- 
are simply plain triangular surfaces. 

The plan shows one-half of cover, or as much as- 
would usually be made from one piece. First divide 
G H of plan into any convenient number of equal 
parts — in this case four — and connect the points thus 
obtained with 0, thus obtaining the base lines of a set 
of right angled triangles whose hypothenuses when 
obtained will give the true distances from the points in 
Gr H to the apex of the cover. 

To construct a diagram of triangles represented! 
by lines in plan, draw the right angle M N P in Fig^ 




Fig. 534.— Diagram of Triangles. 



534, making M X equal to the hight of cover, as 
shown by B D of elevation. Measuring from X, set 
off on X P the length of lines in plan, including J O 
and F. From the points in X P draw lines to M, 
as shown. The line C' M gives the slant hight of 
cover as seen in the end elevation, and M J' the slant 
hight as would be seen in side elevation. The other 
lines give the hypothenuses of triangles, the bases of 
which are shown by the lines in O G H of plan. 

To describe the pattern -proceed as follows : 



310 



Tlie New Metal Worker Pattern Booh. 



Draw the line Q IT, in Fig. 535, in length equal to 
M J' in the diagram of triangles. Through TJ, at right 
angles to Q U, draw V T, making U T and U V each 
equal to J II or J K of Fig. 533, and draw Q T and 
Q V. Then Q V T will be the pattern of one of the 
sides of the pyramid, to which may be added on either 
side the envelope of the portion of a scalene cone shown 
by H G in Fig. 533. It should be here remarked 
that the method above employed of obtaining the 
length Q T produces the same results as that employed 
in the diagram of triangles as shown by the hypoth- 
enuse M 1, which is one side of the adjacent triangle 
forming part of the pattern of the rounded corner. From 
Q of Fig. 535 as center and M 2 of the diagram of tri- 
angles as radius strike a small arc, 2', which arc is to 
be intersected with one struck from T of pattern and 
the distance H 2 of plan as radius. Proceed in this 
manner, using the spaces in H G of plan for the dis- 
tances in T S of pattern, and the lengths of lines drawn 
from M to points 2 to 5 in diagram of triangles for the 
distances across the pattern from Q to the points in 
T S. "With S of pattern as center, and Gr F of plan as 
radius, describe a small arc, R, which intersect with 
one strack from Q of pattern as center, and M C of the 
triangles or B G of the elevation as radius, thus estab- 



lishing the point R of pattern, 
line through the points from S 

Q 



Draw R S, and trace a 
to T, as showD. The 




Fig. 535. — Pattern of Cover. 

other part of pattern, as Q V W P, can be described 
in the same manner, or by duplication. 



PROBLEM 166. 



Pattern for the End of an Oblong: Vessel which is Semicircular at the Top and Rectangular at the Bottom. 



In Fig. 536, A C D E represents the side eleva- 
tion of the article, F H K L the end elevation, and M 
NEPin Fig. 537 the plan. By inspection of these 
it will be seen that the portion represented upon the 




E D 

SIDE. END. 

Fig. 536. — Side and End Elevations of Oblong Vessel with 
Semicircular End. 



end elevation by G L K is simply a flat triangular sur- 
face, while the corners of the vessel, shown by B C D 



of the side view and N R of the plan, are quarters of 
the envelope of an inverted scalene cone. 

To obtain the patterns proceed as follows : Divide 
one-half of the end of the plan into any convenient 





Fig. 537.— Plan of Oblong Vessel with Semicircular End. 

number of equal spaces, all as shown by small figures 
1, 2, 3, 4, etc., in N R. From each of the points thus 



Pattern Problems. 



311 



determined draw lines to the point N", the apex of the 
cone in plan, all as shown in the engraving. Proceed 
next to construct the diagram of triangles shown in 
Fig. 53b, of which the lines just drawn in the plan are 
the bases and B D is the common altitude. Draw A B, 
in length equal to D B of Fig. 536, and at right angles 
to it draw B C, which produce indefinitely. From B 





Fig. 5SS. — Diagram of 
Triangles. 



Fig. SS9.— Pattern of 
End Pieces. 



along B C set off spaces equal to the distances from N 
of the plan to the several points in the boundary line. 
That is, make B 5 of Fig. 53S equal to 1ST 5 of Fig. 
537, and B 4 equal to IN" 4, and so on. And from 
each of the points in B C draw lines to A. Then the 
distances from A to the various points in B C will be 



the distances from the apex of the scalene cone to the 
various points assumed in its base, and will be the 
radii of the arcs shown between D F and E in Fig. 
539. For convenience erect any perpendicular, as A 1 
D of Fig. 539, upon which set off distances equal to the 
length of the lines in the diagram drawn from A, or, 
in other words, make A 1 1 equal to A 1 of the diagram, 
Fig. 538 ; A 1 2 equal to A 2 of the diagram, and so on. 
From A 1 as center, with radius A 1 D, describe the arc 
D E indefinitely. In like manner, from the same 
center, with radius A 1 2, describe a corresponding arc, 
and proceed in this way with each of the other points 
lying in the line A 1 D. 

From any convenient point upon arc 6, as F, draw 
A 1 F, which will represent the side of the pattern cor- 
responding to B D of the side elevation. With the 
dividers set to the space used in stepping off the arc 
N R of the plan, place one foot at the point F of the 
pattern and step from one arc to the next until all the 
arcs are reached, and draw A 1 E. Then A 1 F E will 
be one portion of the required pattern. From E as 
center, with radius E A 1 , describe the arc A 1 G in- 
definitely. Make the chord A' G equal to L K of the 
end elevation, Fig. 536, and draw E G. Then A 1 E G 
will be the pattern of that portion shown by L G K of 
the end view and 1ST R P of the plan. Duplicate the 
part A 1 F E, as shown by G H E, thus completing the 
pattern of the entire end. 



PROBLEM 167. 

Pattern of an Irregular Flaring: Article, both Top and Bottom of which are Round and Parallel, but 
Placed Eccentrically in Plan ; Otherwise the Envelope of the Frustum of a Scalene Cone. 



In Fig. 540 is shown an elevation and plan of the 
article, in which E F Gr H is the plan of the bottom 
and E J K L that of the top, the two being tangent at 
the point E. In Fig. 541 the elevation and a portion 
of the plan are drawn to a larger scale and conven- 
iently located for describing the pattern. 

Since the top and the base of the article are both 
circular and are parallel, the shape of which the pat- 
tern is required becomes a frustum of a scalene cone, 
and lines drawn upon its surface from any set of points 
assumed in the circumference of its base to its apex 
will divide the circumference of the top into similar 



and proportionate spaces. Therefore, the first step is 
to extend the lines of the sides B A and C D until 
they meet at M, the apex. Next divide the plan of 
the base, one-half of which, EH6, only is shown, 
into any convenient number of equal spaces, as shown 
by the small figures. As it is necessary to ascertain 
the distance from each of these points to the apex of 
the cone the simplest method of accomplishing this is 
as follows : From E, the position of the apex in plan, 
as a center, with E 6, E 5, E 4, etc., as radii, describe 
arcs cutting E G. Carry lines vertically from each of 
the points in E G, cutting the base line A D ; thence 



312 



The New Metal Worker Pattern Book. 



carry them toward the apes M, cutting the line of the 
top B C, all as shown. 

With M as center describe arcs from each of the 
points in the base line A D, and extend them indefi- 
nitely in the direction of 0. In the same manner 
draw arcs from the points of intersection in B C, as 
shown. From the apex M draw any line to intersect 
the arc from A or 7 of the base line, as M N, which 
will form one side of the pattern, corresponding to 
B A of the elevation. Set the dividers to the space 




Fig. 640.— Plan and Elevation of Frustum of a Scalene Cone. 



E 6, used in dividing the plan of the base, ana starting 
from N step from one arc to the next, thus laying out 
the stretchout of the base E II G, and at the same 
time locating each point at its proper distance from the 
apex M. A line traced through these points, as N Q 
0, will be the bottom of one-half of the pattern. 
From the points in the line NQO draw lines to M, 
cutting the arcs of corresponding number previously 



drawn from B C ; then a line traced through these 
points of intersection, as shown by B P, will be the top 




Fig. 541. — Method of Obtaining Pattern of a Scalene Cone. 

of the pattern, and P B N Q O will thus be one-half 

the required pattern. 



PROBLEM 168. 

Pattern of a Flaring; Article, the Top of which is Round and the Bottom Oblong- with Semicircular 

Ends.— Two Cases. 



First Case. — In Fig. 542 is shown the elevation 
and plan of the article drawn in proper relation to each 
other, as shown by the lines of projection. In this case 



the top of the article is located centrally with reference 
to the bottom, as shown in the plan. From 0, the 
center of the top, erect tne perpendicular 0, cutting 



Pattern Problems. 



313 



the line of the top in elevation, and from P erect the 
perpendicular P p, cutting the line of the base. Since 
LMK and FHGof the plan are semicircles lying in 
parallel planes, that part of the pattern of the article 
shown by L F H G X M must be one-half the envelope 
of the frustum of a scalene cone. 




Fig. 54S.— Plan, Elevation and Pattern of Flaring Article with Round 
Top and Oblong Base, the Top being Centrally Located, 



To ascertain the apex of the cone, prolong the side 
line C D of the elevation indefinitely in the direction 
of X. Through the points p and draw p 0, which 
produce until it meets C D prolonged in the point X. 
Then X is the apex required. From X drop a per- 
pendicular, cutting the center line of the plan at X 1 , 
thus locating the apex in plan. Divide FHG into 



any convenient number of equal spaces, as shown by 
the small figures. Should lines be drawn from each of 
the points thus obtained to X', they would represent 
the bases of a set of right angled triangles, of which Y 
X is the common altitude, and whose hypothenuses 
will give correct distances from the apex of the cone 
to the various points assumed in the base. 

The simplest method of obtaining these hypothe- 
nuses is as follows: From X 1 as center draw arcs from 
each of the points in F H G, cutting the center line 
E H of the plan. From each point in P H erect a per- 
pendicular to p C, as shown. From the points thus 
obtained in p C carry lines toward the apex X, cutting 
o D, as shown. From X as center strike arcs from 
each of the points in p C indefinitely. Assume any 
point, as G', upon the arc struck from point 1 as the first 
point in the pattern of the base, from which draw a line 
to X. Set the dividers to the space used in stepping 
off the plan, and, commencing at G 1 , step to the second 
arc, and from that point to the third arc, and so on, as 
shown in the engraving. A line traced through these 
points will be the boundary of a lower side of the semi- 
circular end. From each of these points just obtained 
draw a line toward the center X. Place one foot of 
the dividers at X, and, bringing the pencil point suc- 
cessively to the points in D, cut radial lines of cor- 
responding number just drawn. A line traced through 
these points of intersection, as shown by X* L 1 , wit. 
form the upper edge of the pattern of the end piece. 
From the point L 1 , which corresponds to L of the 
plan, as center, with L 1 F 1 as radius, describe the arc 
F' R', and from F 1 as center, with radius equal to F R 
of the plan, intersect it at R 1 , as shown. Draw L 1 R'. 
Then L 1 R 1 F" is the pattern of one of the sides. To 
L 1 R 1 add a duplicate of the end piece already ob- 
tained, all as shown by L 1 R 1 S 1 W, and to S' X 2 add 
a duplicate of the side just obtained, as shown by 
S 1 X a G\ thus completing the pattern. 

Second Case. — This case differs from the first only 
in the fact that the top of the article, being located near 
one end, is drawn concentric with the semicircle of the 
near end. As the result of this condition, that portion 
of its pattern shown by S E R L K X in the plan, Fig. 
513, becomes one-half the envelope of the frustum of 
a right cone, the method of developing which is given 
in Problem 123 of the previous section of this chapter. 

In that portion of the article shown by R F H G 
S X the conditions are exactly the same as in the first 
case. In Fig. 543 corresponding parts have been let- 
tered the same as in Fig. 512, so that the demonstra- 



314 



The New Metal Worker Pattern Book. 



tion given above is equally applicable to either figure. 
In the final make up of the various parts of which the 



a duplicate of L 1 F 1 G 1 N 1 , all as clearly shown in the 
engraving. The seam in this case has been located 




Fig. 543. — Plan, Elevation and Pattern of Flaring Article with Round Top and Oblong Base, the Top being Located 

near One End. 

complete pattern is composed, the part R 1 S 1 N 3 L 1 is upon the line N S of the plan instead of upon N G as 
of course obtained as in Problem 123 instead of being before. 



PROBLEM 169. 

The Patterns of a Flarinj Tub with Tapering: Sides and Semicircular Head, the Head having: More 

Flare than the Sides. 



In Fig. 544, A B C D shows a side elevation of 
the tub, LMNOP the plan at the top, and E F G 
H K the plan at the bottom, an inspection of which 
will show that the head, as shown by H or C D, has 
more flare than the sides, whose flare is shown by A J 
or A B, the flare of the sides and foot being the same. 
Inasmuch as the article is tapering in plan, the conical 
part of the pattern will include a little more than 



a semicircle, as shown. The points showing the 
junction between the straight sides and the conical 
part are to be determined by lines drawn from the cen- 
ters by which the top and bottom were struck, per- 
pendicular to the sides of the article. Therefore lay 
off in the plan T N and T P, drawn from the center T 
of the curved part of the plan of the top of the article, 
perpendicular to the sides M N and L P respectively. 



Pattern Problems. 



31, 



And in like manner from S, the center by which the 
curved part of the bottom of the article is struck, 
draw S G and S K. 

Since the top and bottom of the tub are parallel, 
as shown by the side elevation, and their circles are 
not concentric in xhe plan, it follows that the part P 




Fig. 544. — Plan, Elevation and Pattern of Flaring Tub with Tapering Sides and 

Semicircidar Head. 



N G H K is part of the envelope of a scalene cone. 
To find the apex of this cone, first drop lines from the 
points T and S vertically, cutting respectively the top 
and bottom lines of the elevation, as shown at T 1 and 
S 1 . A line connecting T 1 and S' will give the inclina- 
tion of the axis of the cone in that view, which con- 
tinue indefinitely in the direction of R 1 until it inter- 
sects the side G D continued, as shown at R 1 . Then 



R 1 is the apex of the cone. From R 1 draw R' R verti- 
cally, cutting the center line of the plan at R. Then 
R shows the position of the apex of the cone in the 
plan. As the pattern of the curved portion consists 
of two symmetrical halves when divided by the center 
line of the plan, divide the curve N into any con- 
venient number of equal spaces, as shown by 
the small figures. Lines drawn from each 
of these points to R would represent the 
bases of a series of right angled triangles 
whose common altitude is V R 1 , and whose 
hypothenuses when drawn will represent the 
correct distances from the apex to the vari- 
ous points assumed in the base of the cone. 
The simplest method, however, of meas- 
uring these bases is to place one foot of the 
compasses at the point R, and, bringing the 
pencil point successively to the points in N 
O, draw arcs cutting the center line, as shown 
between T and 0. Now place the blade of 
the T-square parallel to R R' and drop lines 
from each of these points, cutting the line 
A D as shown. From the points obtained 
upon A D draw lines toward the apex R', 
cutting the bottom line of the tub B C. 
These lines drawn from the points in A D 
to R 1 will be the desired hypothenuses and 
may be used in connection with the spaces 
of the plan in developing the envelope of the 
scalene cone. 

Therefore from R 1 as center, and radii 
corresponding to the distance from R 1 to 
the several points in T : D, describe a set of 
arcs indefinitely, as shown. Assume any 
point upon the arc 0, as 1ST', as a starting 
point, from which draw a line to R 1 . With 
the dividers set to the space used in divid- 
ing the plan 1ST 0, place one foot at the point- 
N 1 and swing the other foot around, cutting 
the arc 1. Repeat this operation, cutting 
the arc 2, and so continue to step from arc 
to arc until all the arcs have been reached, 
which will complete the outline of one-half the pattern. 
The operation of stepping from arc to arc can be con- 
tinued, stepping back from arc 5 till arc is reached 
at P 1 , thus completing the top line of the pattern of the 
entire curved portion of the tub. From each of the 
points thus obtained draw lines toward the apex R', as 
shown. Place one foot of the compasses at R 1 , and, 
bringing the pencil point successively to the points in 



316 



The New Metal Worker Pattern Booh. 



the line S' previously obtained, cut radial lines of 
corresponding number in the pattern, as shown from 
G 1 to K'. Lines traced through the several points in 
the two outlines, as shown by G' H' K 1 and N 1 0' P 1 , 



will complete the pattern of the conical part of the tub. 
The patterns of the sides and foot may be obtained 
as described in Problem 74 and as indicated in the 
upper part of the engraving. 



PROBLEM 170. 

The Pattern of a Flaring 1 Article which is Rectangular with Rounded Corners, Having More 

Flare at the Ends than at the Sides. 



In Fig. 545, ABCDEF represents the plan at 
the top of a portion of the flaring article, whose general 
shape is rectangular with rounded corners. GHI J 
K L represents the plan of the bottom of the same, 
showing that the flare at the ends, represented by I C 
or J D, is greater than that of the sides, represented by 
A G. The arc E D of the plan of the top is struck 
from as center, while the arc J K of the bottom is 
struck from N. Since the top and bottom are parallel, 
as shown by P Q and P S of the side elevation, the 
corner JDEK is a portion of the envelope of the 
frustum of a scalene cone. 

To find the apex of the cone, drop lines from O 
and N at right angles to P Q, cutting respectively the 
top and bottom lines in the side view, as shown at T 
and IT, and draw T U and continue the same indefinitely 
in the direction of X. Also continue Q S until it in- 
tersects T U at the point X. Then X will be the apex 
of the cone. From X erect a line vertically, cutting the 
line D M of the plan at M ; then M will show the posi- 
tion of the ajDex of the cone in plan. Divide the arc E 
D into any convenient number of equal spaces, and 
from the points thus obtained draw arcs from M as 
center, cutting D, as shown. From the points in 
Ddrop lines vertically, cutting T Q, the top line of the 
side, otherwise the base of the cone. From the points 
thus obtained in T Q draw lines toward the apex, cut- 
ting U S. 

"With one foot of the compasses set atX, bring the 
pencil point successively to the points in T Q and draw 
arcs indefinitely, as shown. From any convenient 
point upon the arc 0, as E', draw a line to X, forming 
one side of the pattern. Take between the points of 
the dividers a space equal to that used in stepping off 
the plan of cone E D, and, placing one foot at the point 
E', swing the other foot around, cutting the arc 1, from 
which intersection step to the next arc, and so continue 
until all the arcs have been reached at D 1 , from which 



point draw a line to the apex X. Likewise from each of 
the points between E 1 and D 1 draw lines toward the apex 
indefinitely. Finally, with one foot of the compasses at 
X, bring the pencil point to each of the points in U S 




Fig. 54a. — Plan, Elevation and Pattern of Rounded Corner for an 
Article whose Sides and Ends Flare Unequally. 



and draw arcs, cutting radial lines of corresponding 
number in the pattern. Lines traced through the 
points between the points K 1 and J' and E 1 and D 1 will 
complete the pattern of the curved corner. The pat- 
tern for the plain sides can easily be obtained after the 
manner described in Problem 74 and added to that of 
the corner as may be found practicable. 



Pattern Problems. 



317 



PROBLEM 171. 

The Pattern of a Flaring: Article which Corresponds to the Frustum of a Cone whose Base is a True 

Ellipse. 



In Fig;. 546, let G H F E be the elevation of one 
side of the article, L M IT E. the elevation of an end, 
E' R 1 F 1 U' the plan of the article at the base, and T V 
S P the plan at the top. Produce E G and F II of the 
side elevation until they meet in the point I, the apex 
of the cone. 

Divide one-quarter of the plan E 1 R 1 into any con- 




Fig. 546. — Plan, Side and End Elevations of the Frustum of an 
Elliptical Cone. 



venient number of equal parts, as indicated by the 
small figures. From the points thus determined draw 
lines to the center C. These lines will form the bases 
of a series of right angled triangles whose common alti- 
tude is the hight of the cone, and whose hypothenuses 
when drawn will give the true distances from the apex 
to the several points assumed in the base of the cone. 
Therefore at any convenient place draw the straight 
line D A of Fig. 547, in length equal to I E\ Make 
D B equal to I G'. From A and B of Fig. 547 draw 
perpendiculars to D A indefinitely, as shown by A 



and B N". Take the distances C 5, C 4, C 3, etc., of 
the plan and set off corresponding distances from A 
on A 0, as shown by A 5, A 4, A 3, etc. From these 
points in A draw lines to D, cutting B 1ST. These 
lines are also shown in the elevation, but are not 
necessary in that view in obtaining the pattern. From 
D as center describe arcs whose radii are equal to the 
lengths of the several lines just drawn from D to the 
points in A 0. 

From any convenient point in the first arc draw a 




Fig. 547. — Diagram of Triangles and Pattern of Frustum Shown 
in Fig. 546. 



straight line to D, as shown by W D. This will form 
one side of the pattern. From W, as a starting point, 
lay off the stretchout of the plan E' B,' F 1 , etc., using 
the same length of spaces as employed in dividing it, 
stepping from one arc to the next each time, as shown'. 
A line traced through these points will be the outline 
of the base of the pattern, one-half of the entire en- 
velope being shown in the pattern from W to Z, 

From the points in W Z draw lines to D, which 
intersect by arcs drawn with D as center and starting 
from points of corresponding number in B 1ST. A 
line traced through the points of intersection will 
form the upper line of the pattern, as shown. Then 
WXYZ will constitute the pattern of one-half of 
the envelope, to which add a duplicate of itself for 
the complete pattern. 



318 



The New Metal Worker Pattern Booh 

PROBLEM 172. 



The Patterns for a Hip Bath. 



In Fig. 548, let H A L ]S T be the elevation of 
the bath, of which D 1 G E 1 B' is a plan on the line D 
E. Let the half section A ! M C B 3 represent the flare 
which the bath is required to have through its sides 
on a line indicated by A B in elevation. By inspec- 
tion of the elevation it will be seen that three patterns 
are required, which, for the sake of convenience, have 
been numbered in the various representations 1, 2 
and 3. 

Since the plan of piece No. 1 on the line I) B, 




Fig. B4S. — Plan, Elevation and Section of a Hip Bath. 



which is at right angles to its axis A F, is a semicircle, 
as shown by G D 1 B', and since its flare at the side, as 
shown by C B 2 A°, is the same as at B D LT, its pat- 
tern will be a portion of the envelope ^ a right cone. 
Patterns of this class have been treated in the previous 
section of this chapter — to which the reader is re- 
ferred — where, in Problem 143, an exactly similar 
subject has been treated. The operation of obtaining 



this pattern is fully shown in Fig. 549, and, therefore, 
need not be here described. 

Piece No. 2, as shown in Fig. 548, is so drawn as 
to form one-half of the frustum of an elliptical cone. 
As its section at A B (shown at the right) must neces- 
sarily be the same as that of piece No. 1, against 
which it fits, the point F is assumed as the apex of the 
elliptical cone, and consequently the flare at the foot, 
E L, is determined by a continuation of the line drawn 
from the apex through E. Should it be decided to 
have more flare at the foot than that shown by E L, 
the point L may be located at pleasure, and the plan 
of the top, K L l A 1 , be drawn arbitrarily; after which 
its pattern may be developed by means of the alter- 
nating triangles alluded to in the introduction of this 




Fig. 549.— Pattern of Piece No. 1 of Hip Bath. 

section (page 306), examples of which will be found 
further on in this section. 

The plan G E 1 B', from which the dimensions of 
the pattern are to be determined, may be a true ellipse, 
or may be composed of arcs of circles, as shown, ac- 
cording to convenience. Divide one-half the plan G 
E 1 into any convenient number of equal spaces, as 
shown by the small figures, and from each point thus 
obtained draw lines to the center C. To avoid con- 
fusion of lines a separate diagram of triangles is con- 
structed in Fig. 550, in which M G is the hight of the 



Pattern Probleins. 



319 



tub or frustum and M F the hight of the cone. Draw 
M L and C E, each at right angles to M F. Upon C E 
set off from C the lengths of the several lines C 1, 



1534 s it 




Fig. 550.— Pattern of Piece No. 2 of Hip Bath. 

C 2, etc., of the plan. Through each of the points in 
C E draw lines from F, cutting the line M L. From 




Fig. 551. — Diagram of Radii for Pattern of Foot. 

F as center draw arcs indefinitely from each of the 
points in C E and also from the points in M L. 



Through any point upon arc 1 of the lower set, as G, 
draw a line from F and extend it till it cuts arc 1 of 
the upper set at K ; then K G will be one side of the 
pattern. With the dividers set to the space used in 
dividing the plan G E', place one foot at the point G 
of the pattern and step to arc 2, and so continue step- 
ping from one arc to the next till all are reached, as 
at E, and repeat the operation in the reverse order, 
finally reaching B and completing the lower line of 
the pattern. From each of the points in G E B draw 
lines radially from F, cutting arcs of corresponding 




Fig. 552.— Pattern of Foot of Hip Bath. 



number drawn from M L. Lines traced through these 
points of intersection will complete the upper line of 
the pattern. Then G E B A L K will be the required 
pattern of piece No. 2. 

As the plan D 1 G E 1 B 1 has been drawn entirely 
from centers (C, P, S and P 1 ), the pattern of piece 
No. 3 is exactly similar to that described in Problem 
134 of the previous section of this chapter, to which 
the reader is referred. In Fig. 551 is shown a diagram 
for obtaining the radii taken from dimensions given in 
Fig. 54S, while Fig. 552 shows the pattern described 
by means of the radii given in Fig. 551. 



320 



The New Metal Worker Pattern Book. 



PROBLEM 173. 



The Patterns for a Soapmaker's Float. 



In Fig. 553 is shown a perspective view of a soap- 
maker's float. In general characteristics it is very 
similar to piece No. 2 of the hip bath treated in the 
preceding problem. It also resembles the bathtub in 
that its bottom is bulged or raised with the hammer, 
and is therefore not included in the field of accurate 
pattern cutting. The sides are to be considered as 
parts of two cones having elliptical bases, the short 
diameters of which are alike, but the long diameters 
of which vary. 

In Fig. 554 is shown a plan and an inverted eleva- 
tion of the flaring sides, showing in dotted lines the 
completed cones of which the sides form a part. Thus 
L D' M represents one-half the base of an elliptical 
cone of which L M is the short diameter and D 1 K' 
one-half the long diameter. As all sections of a cir- 




J^ig. 553.— Perspective View of Soapmaker's Float. 

cular cone taken parallel to its base are perfect circles, 
so all sections of an elliptical cone parallel to its base 
must be ellipses of like proportions with the base. 
Therefore the plan of the upper base of the frustum 
A' P must be so drawn that a straight line from A' to 
P will be parallel to a straight line joining D 1 and L. 

For the pattern of the portion shown by D A E F 
of the elevation, first produce the line F B of the ele- 
vation in the direction of K indefinitely. In like man- 
ner produce D A of the elevation until it reaches F E 
produced in the point K. Then D K F may be re- 
garded as the elevation of a half cone, of which that part 
of the vessel is a portion, and K F its perpendicular 
hight. Next, divide one-half the plan LD'M into 
any number of equal parts, as shown by the small 
figures 1, 2, 3, etc. Construct the diagram of trian- 



gles, shown in Fig. 555, by drawing the line D K 1 of 
indefinite length, and the line K 1 K at right angles ta 
it, making K 1 K in length equal to F K, Fig. 554. 
Establish the point K a by making the distance K 1 K 3 
equal to E F of Fig. 554. Draw K 2 A parallel to K 1 D. 
From each of the points 2, 3, 4, etc., of the plan draw 
lines to the center K', and set off distances equal to 
these lines upon the line K 1 D of Fig. 555, measuring 
from K 1 toward D. From each of the points thus ob- 
tained draw lines to the point K, cutting A K 3 . "With 
one foot of the compasses in the point K, and the other 



-Oil 




Fig. 554.— Plan and Inverted Elevation of Soapmaker's Float. 

brought successively to the points 1, 2, 3, etc., in the 
line D K' and also to the points in the line A K 2 , de- 
scribe arcs indefinitely. 

Take in the dividers a space equal to the divisions 
in D 1 L of Fig. 554, and, commencing at the point a 
in arc 7 (Fig. 555), step to arc 6 and thence to arc 5, 
and thus continue stepping from one arc to the next 
until the entire stretchout of the half plan has been 
laid off, as shown in Fig. 555. From each of the 
points thus obtained in a d draw lines to K, cutting 
arcs of corresponding number drawn from A K\ Then 



Pattern Problems. 



321 



a line traced through the several points of intersection 
thus obtained, as shown by b c and a d, will be the 
boundary lines of the pattern. 

The pattern for the other end of the article is to 
be, in the main, developed in the same manner as above 
described. One additional condition, however, exists 
in connection with this piece, viz. : To determine the 
dimensions of the cone of which this piece (EBCF, 
Fig. 554) is a part, since the flare at B C is much 
greater than A D, while the flare of both pieces at the 
side is the same as shown by P L of the plan. The 
quarter ellipse P B 1 of the plan being given, and also 
the point C, it becomes necessary to draw from C a 
quarter ellipse which shall be of like proportions with 




01 23 J SU7 



Fig. 555. — Diagram of Triangles and Pattern of Piece 
LD*M of Fig. 554 



P B 1 , which, as remarked above, is a necessary condi- 
tion, both beins: horizontal sections of the same cone. 
To do this proceed as follows : Connect the points P 
and B 1 by means of a straight line. From the point 
C 1 draw a line parallel to P B 1 , and produce it until it 
cuts the line L G 1 , which is a straight line drawn at 
right angles to L M. Then G 1 becomes a point in the 
lower base of the cone corresponding to the point P in 
the upper base. Draw the line G 1 P, and continue it 
until it intersects the long diameter in H 1 . Drop the 
point G 1 vertical from the plan on to the base line D 
of the elevation, as indicated by the point G. Draw a 
line through the points G and E, which produce in- 
definitely in the direction of H. In like manner pro- 



duce the side C B of the cone until it intersects G E 
produced in the point H. Then it will be found that 
the jwint H of the elevation and the point H 1 of the 
plan coincide, as indicated by the line H H 1 . H G of 
the elevation thus represents the axis and H C one 
side of the cone of which the piece E B C G is a part, 
from which it will be seen that this cone is at once 
elliptical and scalene. The operation of developing 
the pattern from this stage forward is the same as in 
the previous case, all as clearly shown in Fig. 556, save 
only in the addition of the triangular piece indicated 
by G E F of the elevation. After completing the other 




I? 'G 



5 i 3 2 1^ 



Fig. 556. — Diagram of Triangles and Pattern of Piece 
L&M of Fig. 554. 



portions of the pattern, this triangular piece is added 
as follows : The distance H 1 L in Fig. 556 is to be set 
off on the line H 1 C in the same manner as the distances 
to the other points — i.e., H 1 L is equal to H 1 L of Fig. 
554. ■ Then L is to be treated in the same manner' as 
the other points, an arc being struck from it, as indi- 
cated in the engraving, by which to determine the 
corresponding point 1/ in the outline of the pattern. 
L 2 G 2 is made equal to L G 1 of the plan, Fig. 554. 
From If draw a line to E. Then E L 2 G 2 will be the 
pattern of the triangular piece indicated in Fig. 554 by 
E F G. It is to be added upon the opposite end of 
the pattern in like manner, as indicated by E 1 G' L 1 . 



322 



Tlie New Metal Worker Pattern Book. 



PROBLEM 174. 

The Envelope of a Frustum of an Elliptical Cone Having: an Irregular Base. 



The form E F K L J shown in Fig. 557, the 
lower line of which is an irregular section through an 
elliptical cone, is introduced here, not as representing 
any particular article or class of forms, but because it 
embodies a principle somewhat different from other 
sections of cones previously given, which may be use- 
ful to the pattern cutter. 

B A C is the side elevation of a cone having 
an elliptical base, one-half of 
which is shown by B 1 H C 
Divide one quarter of the plan, 
as H C, into any convenient 
number of equal parts, as 
shown by the small figures. 
From each of the points of 
division draw lines to the cen- 
ter D 2 , and also erect lines 
cutting the base of the cone 
B C, from which carry them 
toward the apex, cutting the 
lines E F and J L K. The 
first operation will be that of 
obtaining the envelope of the 
complete cone in the same 
manner as described in pre- 
vious problems. 

Construct a diagram of 
triangles, as shown at the 
right, in which A 1 D 1 is equal 
in hight to A D, and at right 
angles to G 1 W and D 1 V, ex- 
tensions respectively of E F 
and B C. Upon D 1 V, measur- 
ing from D 1 , set off the dis- 
tances from jy to the several points in H C, as 
shown. From each of the points thus obtained draw 
lines toward A 1 , cutting G' W. Also from each 
of these points, with A' as center, describe arcs indefi- 
nitely. Take between the points of the dividers a 
space equal to that used in dividing the plan H C, 
and placing one foot upon the arc drawn from point 1 
in the line D' V, step to arc 2, thence to arc 3 and so 
continue till one quarter of the stretchout is completed 
at 7, and, if desirable, continue the operation, taking 
the arcs in reverse order, thus completing the outline 
of one-half the envelope of the cone, as shown in the 



From each of the points in this outline or 
stretchout draw measuring lines toward the center A 1 . 
Place one point of the compasses at point A 1 , and, 
bringing the pencil point successively to the several 
points of intersection on the line G 1 W, cut measuring 
lines of corresponding number, as shown from G 1 to 
G\ Place the T-square parallel to the base B 0, and, 
bringing it successively to the several points of inter- 




im';/. 557. — Patterns for the Frustum of an Elliptical Cone Having an Irregular Base. 



section previously obtained in the curved line L K, 
cut lines of corresponding number drawn from the 
points in D 1 V to A 1 , as shown from X to Y. Finally, 
with one foot of the compasses at A 1 , bring the pencil 
point to each of the points of intersection last obtained 
and cut corresponding measuring lines in the pattern. 
Then lines traced through the points of intersection, aa 
shown from L 1 to L° and from G 1 to G 3 , will complete 
the pattern of one-half the frustum E F K L J. 

Should it be desirable to cut a pattern to fill the 
end J L K of the frustum, as for a bottom in the 
same, it will first be necessary to obtain a correct plan 



Pattern Problems. 



323 



of the line J K L. To accomplish this, set off on the 
lines D 2 7, D" 6, etc., of the plan the lengths of the 
several lines of corresponding number drawn from the 
line G 1 D 1 to the intersections between X and Y, thus 
obtaining the desired line L 3 K 2 . Extend the center 
line B 1 C 1 of the plan, as shown at the right, upon 
which lay off a stretchout of the line L K, taking each 
of the spaces separately as they occur, all as shown by 



Z K 3 , through which draw measuring lines at right 
angles. Place the T-square parallel to B 1 C, and, bring- 
ing it to the several points in the line L 3 K 3 , cut corre- 
sponding measuring lines. Then a line traced through 
the points of intersection, as shown by L 4 K 3 , will be 
the pattern of one-quarter of +he desired piece, which 
may be duplicated as necessary for a half or for the 
entire pattern in one piece. 



PROBLEM 175. 

The Patterns of the Frustum of a Scalene Cone Intersected Obliquely by a Cylinder, their Axes Not 

Lying: in the Same Plane. 

In Fig. 558, let A B C D represent the frustum I A D and B C are the outlines of the slanting sides. 
of an oblique cone, and T S E V U the cylinder that , In Fig. 559 E F G H shows the plan of the frustum 




Fig. 558. — Front Elevation of the Frustum of a Scalene Cone 
Intersected Obliquely by a Cylinder. 

joins the same at the angle indicated. The view here 
given of the frustum is that of its vertical side, so that 



SECTIONS 



Fig. 559.— Elevation, Plan and Sections of the Frustum of a 
Scalene Cone Intersected Obliquely by a Cylinder. 

at its base and K I J G the plan of the top, from 
which the side elevation is projected at the left, D C 



324: 



Tlie New Metal Worker Pattern Book. 



being the base and A D the vertical side. The inter- 
secting cylinder is indicated by F L M G, and its pro- 
file by NOPQ. The diameter of cylinder is the 
same as that of top of frustum. 

Divide the profile NOPQ into any convenient 
number of equal parts, and from the points thus ob- 
tained carry lines parallel with G M, cutting I J G 
and F G of plan, as shown. As the points in the 
profile of the cylinder lie in four vertical planes, indi- 



shown "With b" and d" as centers, strike the arcs 
G b' and G d', thus forming sections of the cone in 
plan corresponding with a b and c d of elevation. The 
four vertical sections above referred to are shown 
below the plan by I' F', e f, J' g' and h' i'. To avoid 
a confusion of lines, the method of obtaining the shapes 
is shown separately in Figs. 560 to 563, in which the 
reference letters are the same as in Fig. 559. 

To obtain the shape of section on line I F in 




b I 

. I 
\ I 



D' 



¥' 




D' 



/' 





Fig. 560. Fig. 561. Fig. 562. 1 

Sections of the Frustum of Scalene Cone Corresponding to Divisions in Profile of Cylinder in Fig. 559. 



cated by the lines 7, 8 6, 1 5, and 2 4-, it will be nec- 
essary, before their intersection can be obtained, to 
construct four vertical sections through the cone upon 
the lines I F, e /, J g, and h i. The point 3 requires 
no section; it being flush with the vertical side of the 
cone, must intersect somewhere on the line A D. To 
obtain the desired sections divide A D of elevation 
into any convenient number of equal parts, and from 
the points thus obtained erect lines parallel to the base 
D C, cutting B C. From the points in B C carry lines 
parallel with A G, cutting the center line E G, as 



Fig. 560, extend E G, as indicated by I' D', which 
make equal to A D of the elevation, with its points of 
division a and c. From the points in I' D' erect the 
perpendiculars a b, c d and D' F'. With the T square 
placed parallel with I' D', drop lines from the points 
in I F, cutting similar lines drawn from I' D'. A line 
traced through the points of intersection, as shown by 
I' F', will give the required shape. The sections 
shown in Figs. 561, 562 and 563 are obtained in a 
similar manner. 

Having obtained these sections of the cone by the 



Pattern Problems. 



325 



above method, arrange them as shown below the plan 
m Fig. 559. An inspection of the plan and profile 
will show that a line drawn from of profile will cut 
section I F, line S 6 of profile will cut section ef, line 
1ST P of profile will cut section J g, line 2 4 will cut 
section h i, and a line from Q of profile will cut the 
vertical side represented by Gr. 

In connection with the sections in Fig. 559 draw 
an elevation of cylinder, as shown by S R V TJ, oppo- 
site the end of which draw a profile, as indicated by 




Fig. 564. — Method of Obtaining Pattern of Cylinder Shown in Figs. 553 and 559. 



N' Q' P' 0', commencing the divisions at the point 
N'. From the several points in the profile N' 0' P' Q' 
carry lines parallel with TJ V against the several pro- 
files I' F', e' f, J' g' and li i' as described above and 
as indicated by the small figures 1 to 8. A line traced 
through these points of intersection will give the miter 
line. A duplicate of this part of Fig. 559 is presented 
in Fig. 564: for the purpose of avoiding a confusion of 
lines. The miter line drawn through the intersecting 
lines is indicated by S T TJ. 

Having now the profile of the cylinder and the 



miter line, all as shown, the pattern of the cylinder is 
obtained in accordance with the principles given in 
numerous examples in Section 1 of this Chapter, and 
as clearly shown in Fig. 564. 

The method of obtaining the envelope of the 
frustum, and the opening in the side of the same to fit 
against the end of the cylinder just obtained, is shown 
in Fig. 565. The simple envelope of the frustum is 
obtained exactly as described in Problem 167, as will 
be seen by a comparison of Figs. 565 and 541. To 
obtain the opening in its side, however, involves an 
operation similar to that given in the problem immedi- 
ately preceding. ABCD of Fig. 565 represents a 
side elevation of the frustum, as shown by the same 
letters in Fig. 559, and the vertical lines drawn 
through the same, designated by the small figures at 
the bottom, are the lines of the vertical sections ob- 
tained in Figs. 560—563, and correspond in numbers 
to the divisions in the profile in Fig. 559. To obtain 
the elevation of the opening, set off on each of these 
section lines the hights of the points of intersection 
occurring on corresponding sections as they appear in 
Fig. 564. Thus upon line designated at the bottom 
by 2 4, set off the vertical hights of points 2 and 4 on 
section h' i\ Fig. 564, which section corresponds to line 
2 4 of the profile, as shown in Fig. 559. In the same 
manner set off online 1 5 the verti- 
cal hights of the points 1 and 5 on 
section J' g . Obtain also points 8 
and 6 from section e f and point 7 
from section I' F', all as shown by 
the small figures. A line traced 
through these points will give the 
elevation of the opening. The out- 
line of the opening has been shown 
in the plan, but its development is 
not necessary to the subsequent 
work of obtaining the pattern. 

Divide the plan of the base of 
the frustum GrFE into any convenient number of equal 
spaces, as indicated by the small letters. As an accu- 
rate elevation of the opening has now been obtained, this 
operation can be conducted without reference to any of 
the points previously used in obtaining the line of the 
opening. Therefore letters have been used in the divis- 
ions of Gr F E instead of figures so that no confusion 
may arise. From each of these points of division draw 
lines to Gr, which represents the plan of the apex of the 
cone. Also from so many of these points from which 
lines will cut the line of the opening, as a to /, erect 



326 



'Hie New Metal Worker Pattern Book. 



lines verticany, cutting the base line D C, as shown by 
corresponding letters. From these points draw lines 
toward the apex of the cone X, cutting the line of the 
opening in the elevation, as shown but not lettered. 

Proceed now to construct the diagram of triangles 
shown at the right, in which X 1 D 1 is equal to and 
parallel with X D, and in which A 1 B 1 and D 1 C are 
drawn in continuation of A B and D C, as shown. 



dividers at D 2 , step to arc b, thence to arc c, etc., till 
arc i is reached at C\ A line traced through these 
points will give the lower outline of the half of the 
envelope of the frustum which is pierced by the cylin- 
der. From each of these points also draw lines toward 
X 1 , which intersect by arcs of corresponding number 
drawn with X 1 as center from the line A 1 B 1 , thus ob- 
taining the upper line of the envelope. 




Fig. 565.— Method of Obtaining Opening in Side of Cone to Fit End of Cylinder. 



Upon D" C, measuring from D', set off the several 
lengths G 5, G c, etc., of the plan, as shown by cor- 
responding letters, and from the points thus obtained 
draw lines to X', cutting the line A 1 B 1 . From X 1 as 
center draw arcs indefinitely from each of the points 
in D 1 C 1 . Zrom any convenient point upon arc a, as 
D 2 , draw a line to X 1 , which will form one side of the 
pattern of the desired envelope. Take between the 
points of the dividers a space equal to that used in 
dividing the plan G F E, and, placing one foot of the 



From each of the points where the lines b b, c c, 
d d, etc., of the elevation cross the line of the opening 
project lines horizontally, cutting hypothenuses of 
corresponding letter in the diagram of triangles, all as 
shown by a\ b' b\ c 1 c\ etc. "With one foot of the com- 
passes at X 1 , bring the pencil point successively to the 
points a 1 , 5', b% etc., and draw arcs cutting radial lines 
in the pattern of corresponding letter. Then a line 
traced through the points thus obtained will be the re- 
quired shape of the opening in the pattern. 



Pattern Problems. 

PROBLEM 176. 

The Pattern for a Chimney Top. 



327 



In Fig. 566 are shown the side and end elevations 
and the plan of a chimney top. A B C I> of the plan 
represents the size of the article at the bottom to fit 
the chimney, and E F G H is the size of the opening 




from the points thus obtained draw lines to C. The 
next step is to construct a diagram of triangles of 
which the lines just drawn are the bases and of which 
the hight of the article is the altitude. Assuming G 
as the base line of this diagram, place 
one foot of the compasses at C, and, 
bringing the pencil point to the vari- 
ous points in F G, strike arcs cut- 
ting C, as shown. At right angles 
to C erect C Q, equal in hight to 
J H' of Fig. 566, and from Q draw 
lines to the several points in C. 
These hypothenuses will then repre- 
sent the true distances from C to the 
points in F G. From Q as center, 
and radii equal to the several hy- 
pothenuses, strike arcs indefinitely, 
as shown to the left. From any con- 
venient point on arc 0, as G', draw a 
line to Q, which will form one side of 
the pattern of the rounded corner. Set the dividers to 
the space used in stepping off the arc F G, and, com- 



PLAN 
Fig, 566. — Plan and Elevations of Chimney Top. 

in the top to fit a pipe or extension. An inspection 
of the drawings will show that the article consists of 
two flat triangular sides, of which AHD is the plan 
and A 1 H 1 D 1 the elevation, two similar triangles, D G 
C, forming the ends, and four corner pieces, of which 
F C G is a plan, which are portions of an oblique cone. 
Further inspection of the plan will also show that the 
2ntire article consists of four symmetrical quarters, 
therefore in Fig. 567 is shown a quarter plan of the 
same in which corresponding points are lettered the 
same as -in Fig. 566, from which the pattern for one- 
quarter of the article is obtained. 

The quarter circle F G is the quarter plan of an 
oblique cone, of which C is the apex ; therefore divide 
F G into any convenient number of equal parts, and 




Fig. 567.— One-Quarter Plan and Pattern of Chimney Top. 

mencing at the point G 1 , step to arc 1 and from that 
point to arc 2 and so on, reaching the last arc in the 



32a 



lite New Metal Worker Pattern Booh; 



point F 1 . Trace a line through these points, as shown 
from F 1 to G 1 , and draw F 1 Q, which will complete the 
pattern of the corner piece. 

From C set off on C the distances F and L G, as 
shown by M and N. Draw lines from these points to 
Q, then M Q and N Q will represent the true distances 
shown by F and G L of the plan or J H 1 and L G 2 of 
the elevations in Fig. 566. 

"With Q as center, and C L as radius, describe an 
arc, L 1 , and from G 1 as center, with radius equal to N 
Q, intersect the arc, as shown, thus establishing the 
point L 1 . Draw G" L' and L 1 Q. In a similar manner, 
with F 1 as center, and Q M as radius, describe the arc 
0', and from Q as center, with a radius equal to C O 
of the plan, intersect the arc at the point O 1 . Draw Q 
0' and 0" F 1 ; then F' O' Q L 1 G" will form the pattern 
for one complete quarter of the chimney top. A du- 
plicate of this pattern may be added to it if desired, 




Fig. 568.— One-Half Pattern of Chimney Top. 

joining the two upon the line F 1 0, thus forming a 
pattern for one half, as shown in Fig. 568. 



PROBLEM 177. 

The Pattern of an Article with Rectangular Base and Round Top. 



In Fig. 569 are shown the plan and elevations of 
an article in which the conditions are exactly the same 
as in the preceding problem. The article here shown 
differs from that shown in Fig. 566 only in the fact 
that the diameter of the round end or top is greater than 
the width of the base, while in Fig. 566 it is less, but 
the method of obtaining the pattern is exactly the same. 

In this case, as in the preceding one, the article 
consists of four flat triangular pieces (two ends and two 
sides) and four equal rounded corners, each of which is 
a quarter of an oblique cone. As the eivtire envelope 
consists of four symmetrical quarters, one-quarter of 
the plan OPN J has been reproduced in Fig. 570 from 
which to obtain the patterns in the simplest manner. 

Divide J I of plan into any convenient number of 
equal parts, and from the points thus obtained draw 
lines to N, which represents the apex of an inverted 
oblique cone. The object is to construct triangles 
whose altitudes will be equal to the straight hight of the 
article, and whose bases will be equal to the length of 
lines in I J N of plan, and whose hypothenuses will give 
the distance from points in I J of top to N in the base. 



To construct this diagram, proceed as follows: 
From 1ST of Fig. 570 as center, and radii equal to the 
lengths of the several lines drawn to 1ST, describe arcs, 
cutting any straight line, as N W. From N draw N n 
at right angles to N W, which make equal to the straight 
hight of the article, and from the points inN¥ draw 
lines to n. With n as center, and the distances from 1ST 
to points in N W as radii, strike arcs as shown. From 
any point, as i, on arc 1, draw a line to n. Set the 
dividers to the space used in stepping off I J of plan, 
and, commencing at i, step from arc to arc, as indicated 
by the small figures, reaching the last in the point 7 
or/. Draw/n, thus completing the pattern for part 
of article indicated in plan by I 1ST J. From NodN 
W set off the distances Q J and I P, as shown by the 
points q' and f. Then n q' and n f will represent re- 
spectively the altitudes of the flat triangular pieces 
forming the sides and the ends of the article. With 
N Q of plan as radius, and n of pattern as center, strike 
a small arc (f), which intersect with one struck from / 
of pattern as center, and n q' of diagram as radius, thus- 
establishing the point q of pattern. Draw n q and qj„ 



Pattern Problems. 



820 



"With PNof plan as radius, and n of pattern as center, 
strike a small arc, which intersect with one struck from 
i of pattern as center, and n t' of the diagram as radius, 

B E' B' 



In Fig. 571 ipqjls, a duplicate of the pattern 
indicated by the same letters in Fig. 570. Below this 
the pattern is duplicated once, and above twice, each 




E" 



N' M 

Fig. 569.— Plan and Elevations of an Article with Rectangular Base and Round Top. 




Fig. 570. — One-Quarter Plan and Pattern of Article Shown in Fig. 569. 




Fig. 571. — The Entire Pattern of Article shown 
in Fig. 569 in One Piece. 



thus establishing point p of pattern. Draw ip&ndpn, 
as shown, thus completing the quarter pattern. 



alternate pattern being reversed, thus completing the 
entire pattern in one piece. 



330 



Tl\e New Metal Worker Pattern Book. 



PROBLEM 178. 

Pattern for an Article Forming: a Transition from a Rectangular Base to an Elliptical Top. 



In Fig. 572, ABB'A' of the plan shows the 
rectangular base and CEDF the elliptical top of an 
article, the sides of which are required to form a tran- 
sition between the two outlines. A" C D' B" is an 



obtaining the pattern has, however, been employed, 
not because it is better but for the sake of variety, 
leaving the reader to judge which method is the more 
available in any given case. Divide E D into any 

convenient number o f 
equal spaces, as shown by 
the small figures, and from 
the points thus obtained 
draw lines to B. These 
lines will form the bases 
of a series of triangles 
whose common altitude is 
equal to the hight of the 
article, X Y, and whose 
hypothenuses when ob- 
tained will be the real dis- 
tances from B in the base 




Fig. 57S. — Plan and Elevation of Transition Piece. 

end elevation of the same, showing its vertical hight X 
Y. An inspection of the plan will show that the arti- 
cle consists of four symmetrical quarters, and that that 
part of either quarter lying between the curved outline 
of the top and the extreme angle of the base, as the 
part E D B, is a portion of the envelope of an oblique 
elliptical cone, of which E D is the base and B the 
apex. 

The conditions here given are exactly the same as 
in the two preceding problems; a different method of 



Fig. 573.— Pattern for One-Quarter of Article Shown in Fig. 572. 

to the points assumed in the curve of the top. To con- 
struct such a diagram of triangles, first draw any line, 
as L M, and from M lay off the distances shown by solid 
lines in plan, thus making M 1 equal to B 1, M 2 equal 
to B 2, etc. At right angles to L M draw M N, in 
hight equal to the straight hight of the article, as shown 
by X Y of elevation, and connect the points in M L with 
N. Also setoff the distance D divom M, and draw X 
d. If E d was different in length from D d, this distance 
would be set off from M and a line drawn to N. 



Pattern Problems. 



331 



To develop the pattern first draw any line, as E B 
of Fig. 573, equal in length to N 1 of the diagram. 
With B of pattern as center, and N 2 of the diagram 
of triangles as radius, describe a small arc, which in- 
tersect with another arc struck from E of pattern as 
center and E 2 of plan as radius, thus establishing 
point 2 of pattern. Proceed in this manner, using the 
distance between points in plan for the distance be- 
tween similar points in pattern, and the hypothenuses 
of the triangles in the diagram in Fig. 572 for the dis- 
tances to be set off from B of pattern on lines of simi- 
lar number. Through the points thus obtained trace a 
line, as shown by E D. With B of pattern as center, 
and B d of plan as radius, strike a small arc, which in- 



tersect with another struck from D of patterns as cen- 
ter, and N" d of diagram as radius, thus establishing the 
point d of the pattern. Draw D d and d B. With B 
of the pattern as center, and B e of the plan as radius, 
strike a small arc, which intersect with another struck 
from E as center, with a radius equal to N d of the 
diagram of triangles. Draw E e and e B ; then E e B 
d D will be the pattern for one-quarter of the article. 

In performing the work of development of the pat- 
tern it will be found convenient as well as more accu- 
rate to use two pairs of compasses, one of which should 
remain set to the space used in dividing the curve E D of 
the plan, while the other may be changed to the varying 
lengths of the hypothenuses in the diagram of triangles. 



PROBLEM 179. 

Pattern for an Article Forming: a Transition from a Rectangular Base to a Round Top, the Top Not 

Being Centrally Placed Over the Base. 



In Fig. 574, F G H J of the plan represents the 
bottom of the article and A B D E the top. Below 




J' 



FRONT ELEVATION 

Fig. 574. — Elevations and Plan of an Irregular Transition Piece. 



the plan is projected a front elevation and at the right 
a side elevation, like points in all the views being 
lettered the same. An inspection of 
the drawing will show that each side 
of the article consists of a triangular 
piece whose base is a side of the rect- 
angle and whose vertex lies at a point 
in the circle of the top, the four 
vertices marking the division of the 
circle into quarters, and four quarters 
of inverted oblique cones whose bases 
are the quarter circles of the top and 
whose apices lie at the corners of the 
rectangle. A comparison between 
this figure and the one shown in 
Problem 177 will show that the con- 
ditions existing in either one of the 
corner pieces in this case are exactly 
the same as in the former problem, 
but that while in Problem 177 the 
four corners are alike, in the present 
instance the four corners are all differ- 
ent, and that therefore the pattern for 
each corner piece, as well as that for 
each of the flat sides, must be ob- 
tained at a separate operation, all being 
finally united into one pattern. 

Divide the plan of the top A B 
D E into any convenient number of 
equal spaces in such a manner that 



532 



Tlie New Metal Worker Pattern Book. 



each quarter of the circle shall contain the same 
number of spaces and from the points of division in 
each quarter draw lines to the adjacent corner of the 
rectangle of the base, all as shown in Fig. 575. Thus 
lines from the points in E D are drawn to H and 
lines from points in D B are drawn to G, etc. 

The next operation will be to construct the four 
diagrams of triangles (one for each corner piece) 
shown in Fig. 576, of which these lines are the bases. 
Accordingly lay off at any convenient place the line 
H L, Fig. 576, equal to the straight hight of article, 
as shown by J' X in Fig. 574. From the point H, 
and at right angles to L H, draw the line H M, and, 
measuring from H, set off the length of lines in E D 



IB" 



12 / 
/ \ 


i/Y 


\ w 


/ / 
/ 


F 




A 
l\ 14 

1 


/ X- 16 


C D 5 f 

\ 
\ 

PLAN 4 X | 

X ^ 



Fig. 575. — Plan of Irregular Transition Piece with Surface 
Divided into Triangles. 



H of the plan, Fig. 575. Thus H 1 is made equal to 
II E of the plan, H 2 is made equal to the distance 
H 2 of the plan, and H 3 is equal to distance H 3 of 
the plan, etc. From the points thus established in 
M H draw lines to L, as shown. Then the hypoth- 
enuses L 1, L 2, L 3, etc., will correspond to the width 
of the pattern, measured between points in E D of top 
and H in the base. 

The triangles for the corner piece D Gr B are con- 
structed in the same general manner. N Gr corre- 
sponds to the hight J' X of the elevation. Gr is 
drawn at right angles to N Gr, on O Gr are set off the 
lengths of lines in D Gr B of the plan, and from the 
points thus obtained lines are drawn to N. Thus Gr 5 



of the diagram is equal to Gr 5 of the plan, Gr 6 of the 
diagram is equal to Gr 6 of the plan, etc. The tri- 
angles in P Q F correspond with the lines in A F B of 
the plan, as do those in S B J with the lines A 
J E in the plan. Before commencing to desmbe the 
pattern the seam or joint may, for convenience, be 
located at K E of the plan. The real length of the line 
K E of the plan is given by H" E" in the side eleva- 
tion, Fig. 574, or the distance E K can be set off, as 
shown by II K in Fig. 576. The dotted line K L 
will then be the distance from K in the base to E in 
the top. 

As it is necessary in obtaining the pattern for the 
entire envelope that the patterns of the parts shall sue- 





L 






M// /, 


^7/ 1 
/// i 

// i 



5 1 4 2 3 




10 



11 



,3.2 'f ,3 ,41 ;| 

Fig. 576. — Diagrams of Triangles Obtained from Fig. 575. 

ceed one another in the order in which they occur in 
the plan, the method of development here adopted is 
that of constructing separately each small triangle, as 
in the preceding problem, instead of by means of a 
number of arcs, as in Problem 177 and others preced- 
ing it. To begin, then, with the pattern of the part 
corresponding to F B Gr of the plan. ' The length F Gr 
of the pattern, in Fig. 577, is established by the length 
F Gr of the plan in Fig. 575. With F of pattern as 
center, and P 9 of Fig. 576 as radius, describe an arc, 
B, -which intersect with one struck from Gr of the pat- 
tern as center, and 1ST 9 of Fig. 576 as radius, thus es- 
tablishing the point B of the pattern. Then F B Gr ie 



Pattern Problems. 



333 



the pattern for that part of the article shown by F B 
G of the plan. From G of pattern as center, with 
radii corresponding to the hypothenuses of the tri- 
angles shown in N" G of Fig. 576, strike the arcs 
shown. Thus G S of the pattern is equal to N 8, G 7 
of the pattern is equal to 1ST 7, etc. With the dividers 
set to the same space used in stepping off the plan, 
with B or 9 of the pattern as center, strike a small arc 
intersecting arc S previously drawn, thus locating the 
point 8. From S as center intersect arc 7, and so 
continue, locating the points 6 and 5. Through the 
points thus obtained can be traced the line B D. Then 



With the dividers set to the same space as was used in 
stepping off the plan, and commencing at 5, intersect 
each succeeding arc from the point obtained in the one 
before it, as shown by the figures 4, 3, 2, 1. Trace a 
line through the points thus obtained, and connect W 
H, as shown. Then E' D H is the pattern for that part 
of the article shown on plan by E D H. With II of 
pattern as center, and H K of plan as radius, describe 
a small arc, which intersect with one struck from E' 
of pattern as center, and L K of Fig. 576, or what is 
the same, E" H" of Fig. 574, as radius, thus establish- 
ing the point K' of the pattern. Connect H K' and 




Fig. 577.— Pattern tf Transition Piece Shoivn in Fig. 574. 



GBDis the pattern for that part of the article shown 
on the plan by G B D. 

With G of pattern as center, and G H of plan as 
radius, strike a small arc, H, which intersect with one 
struck from D of pattern as center and L M of Fig. 
576 as radius, thus establishing the point H of pattern. 
Connect G H and H D, as shown. Then G E D is the 
pattern for that part of the article shown in plan by G 
H D. With H of pattern as center, and the hypoth- 
enuses of triangles in M L H of Fig. 576 as radii, strike 
arcs, as shown, making H 4, H 3, H 2, H 1 of pattern 
equal to L 4, L 3. L 2, L 1 of the diagram of triangles. 



K' E', as shown, which gives the pattern for that part 
of the article shown on plan by H K E. 

The radii for striking the arcs in A F B of the 
pattern are found in P F of Fig. 576. The length 
F J of pattern is established by the length F J of the 
plan. The radii for striking the arcs in A J E of pat- 
tern are found in S R J of Fig. 576. J K of the pat- 
tern corresponds with J K of the plan, and E K of the 
pattern corresponds with L K of Fig. 576. Thus E A 
BDE' of the pattern is the stretchout of E A B D of 
the plan of the top, as K J F G H K' of the pattern is 
the stretchout of K J F G H of the plan of the base. 



PROBLEM 180. 

The Pattern for a Collar Round at the Top and Square at the Bottom, to Fit Around a Pipe Passing; 

through an Inclined Roof. 



Let A B D C of Fig. 578 represent the side eleva- 
tion of the pipe and C D E F the side view of the col- 
lar, fitting against the pitch of the roof shown by G H. 



Construct a plan below the elevation, as shown, making 
J K M L the plan view of the pipe and NOPE the 
plan view of the collar on a horizontal line, giving the 



334 



Tfte New Metal Worker Pattern Book. 



collar an equal projection at the bottom on the four 
sides, as shown. Through the center point X in plan 
draw a line parallel to N E, intersecting the circle at 
K and L ; likewise through the center X, and 
parallel to N, intersect the circle at J and M. From 
J and K draw lines to the corner N ; likewise from J 
and L, L and M, and M and K, draw lines to the cor- 
ners R, P and 0. It will be seen that by this opera- 



the bases of a series of right angled triangles, whose 
hypothenuses will give the correct distances across the 
pattern of the collar. To construct these triangles 
proceed as follows : Upon C Y extended assume any 
point, as S, at which erect the perpendicular ST, equal 
in hight to the cone CYF, as shown by the dotted 
line from F. From S on S C set off the lengths of the 
several lines in K N J of the plan, as shown by 1', 2', 




Fig. 578. — Plan and Side Elevation of a Collar to Fit Around a Pipe Passing through an Inclined Roof. 



tion the collar has been divided in such a manner that 
the four corner pieces are portions of oblique cones 
whose apices lay at the corners of the collar, while the 
side pieces between are simply flat triangular pieces of 
metal. The dotted lines connecting the plan with the 
elevation show corresponding points in the two views. 
Divide the quarter circles K J and J L into any 
convenient number of equal spaces, as shown by the 
small figures, and from points on each draw lines to 
the corners N and R. Then will these lines represent 



3', etc., and from these points draw lines to T. In 
the same manner construct the diagram of triangles 
shown at the right. At U upon the line Y D extended 
erect the perpendicular U V, equal in hight to the cone 
Y D E, as shown by the dotted line E V. From U 
on U D set off the lengths of the lines in J R L and 
draw the hypothenuses, as shown. 

To develop the pattern, first draw any horizontal 
line, as A A 1 of Fig. 579, equal in length to P R in 
plan of Fig. 578. With A and A 1 as centers, and the 



Pattern Problems. 



33; 



hypothenuse V 9' of Fig. 578 as radius, describe arcs 
intersecting each other at 9. Now, with 9 of the pat- 
tern as center, and 9 8 of the plan as radius, describe 




A A 

Fig. 579.— Pattern of Collar Shown in Fig. 578. 

the arc 8 ; then with V 8' of Fig. 578 as radius, and A 
of the pattern as center, describe an arc, intersecting 
the arc previously drawn, thus establishing the point 8. 
Proceed in this manner, using alternately first the 
divisions on the quarter circle L J in plan, then the 
hypothenuses of the triangles whose bases are shown by 
the lines in JLK, until the point 5 in pattern has 



been obtained. Draw a line from 5 to A in Fig. 579. 
Then with A as center, and E F in side elevation, 
Fig. 578, as radius, describe an arc, shown at C of 
Fig. 579, and with 5 of the pat- 
tern as center, and the hypotlie- 
nuse T 5' of Fig. 578 as radius,' 
describe an arc intersecting the 
previous arc at C. Draw a line 
from 5 to C. Now proceed as 
above described, using alternately 
first the spaces on the quarter 
circle in J K in plan, then the 
hypothenuses of the triangles 
whose bases are shown in J K N 
in plan, until the point 1 in pat- 
tern has been obtained. Then 
with C of the pattern as center, 
and W N of the plan as radius, 
describe an arc, as shown at D, 
and with F C in side elevation as 
radius, and 1 of the pattern as 
center, describe an arc, inter- 
secting the arc previously de- 
scribed at D. Draw the lines 1 
D, D C, C A, and through the 
intersections of the arcs trace a line, shown from 
1 to 9 on pattern. This will complete one-half the 
pattern. The entire pattern may be completed by 
duplicating the part 1 5 9 A C D and adding the same 
to that already obtained in such a manner that the 
side 9 A will coincide with 9 A', as shown by 9 5' 1' 
D' C" A 1 . 



PROBLEM 181. 



The Pattern for a Flaring: Article Round at the Base and Square at the Top. 



The shape shown in Fig. 580 differs from that 
treated in Problem 176 principally in the fact that the 
round end is larger than the rectangular end instead of 
smaller as in Fig. 56G ; the conditions involved are, 
however, exactly the same as in the other problem 
and consequently the method of obtaining the pattern 
must be similar. F G H J, in Fig. 580, represents 
the plan of the base, K L M N that of the top, and 
A B C E the elevation of a side of the article. 



Through 0, the center of the circle of the base, draw 
the diameters G J and F H parallel to the sides of the 
top. From the four points thus obtained in the cir- 
cumference of the base draw lines to the angles of the 
top, as shown by G M and HM,HN and J N, etc. 
It will be seen from this that the envelope of the ar- 
ticle consists of four flat triangles, of which L6M is 
a plan and B D C the elevation, and four rounded 
corners, either one of which, as J N H, is a portion of 



The Nexv Metal Worker Pattern Booh. 



an oblique cone of which J H is the base and N the 
apex. 



g ELEVATION Q 




PLAN 
Fig. 580.— Plan and Elevation of a Flaring Article, Round at 
the Base and Square at the Top. 

To obtain the pattern first divide the quarter plan 
of base J H into any convenient number of parts, as 
indicated by the small figures, and con- 
nect these points with 1ST, as shown. 
To obtain the distance from points in 
J H of base to N of top it will be 
necessary to construct the diagram of 
triangles shown in Fig. 581. Draw any 
line, as E P, in length equal to the 
hight of the article, as shown by S C in 
Fig. 580. At right angles to R P draw 
P Q, and on P Q lay off the lengths of 
lines in J H N. Thus make P 1 equal 
to N 1 of the plan, P 2 equal to N 2, 
etc. Connect the points in P Q with 
R. The hypothenuses thus obtained 



give the true distances from the points in the base to 
N in the top. 

From any convenient point, as N in Fig. 582, as 
center, with radius R 1 of Fig. 581, describe an arc, 
as shown by 1 7. In like manner, with radii R 2, 
R 3 and R 4 of Fig. 581, describe arcs, as shown. 
Draw a straight line from 1ST to any convenient point 
upon the arc 1 7, as shown by 1ST H. Set the dividers 
to the space used in stepping off the plan of the 
base and, starting with H, lay off the stretchout, 




Fig. 581. — Diagram of Triangles Used in Obtaining Pattern of 
Article Shown in Fig. 580. 



stepping from arc to arc, as shown. A line traced 
through these points will form the pattern for as much 
of the article as shown by J 1ST H of the plan. With 
N" of pattern as center, and 1ST K of plan or B C of ele- 
vation as radius, describe a small arc, K, which inter- 
sect with an arc struck from J of pattern as center and 
J 1ST as radius. Connect J K and K N, which com- 
pletes the pattern for J K N of the plan. J K F of 
pattern is the same as J N H, and can be obtained in 
the same manner, or by any convenient means of du- 
plication. With N as center, and N W of plan as 
radius, describe a small arc, which intersect with one 




Fig. 582.— One-Half of Pattern of Article Shown in Fig. 580. 



Pattern Problems. 



337 



struck from H as center, and E C of elevation as F K V of pattern is obtained in a similar manner, 
radius. Connect H W and W N, thus producing the Then TKNWHJF is the pattern for one-half of 
part of pattern corresponding to N W H of the plan. article. 



PROBLEM 182. 

Pattern for an Article Rectangular at One End and Round at the Other, the Plane of the Round End 

Not Being: Parallel to that of the Rectangular End. 



In Fig. 583 are shown front and side views and 
plan of an article forming a transition between a rect- 
angular pipe at one end and a round pipe at the other, 
and forming; at the same time an angle between the 




Fig. 58S. — Front and Side Views and Plan of an Article Forming a Transition Between 
a Rectangular Pipe and a Round Pipe, at an Angle. 



two pipes. A E F C of the front view shows the size 
of the rectangular pipe, while 6 B II D shows the 
opening to receive the round pipe. In the side view 
a c shows the vertical rectangular end, and b d shows 
the angle at which the round end is placed, bid being 
a half profile of the round end. As will be seen by an 
inspection of the front view, each quarter of the circu- 
lar opening is treated as the base of a portion of a 
scalene cone whose apex is in the adjacent angle of the 
rectangle, the intermediate surfaces being flat triangu- 



lar pieces. Thus B & and G D are the quarter bases 
of scalene cones whose apices are respectively at A 
and C ; A B E and CDF are triangles whose altitudes 
or profiles are shown respectively by a b and e d of the 
side view ; and A G C is a triangle 
whose profile appears at u in the 
plan. 

Divide each quarter of the pro- 
file I I and I d of Fig. 583 into any 
number of equal spaces, as shown 
by the small figures ; also draw a 
duplicate of this half profile in 
proper relation to the plan, as shown 
by m g x, which divide as before, 
numbering the points in each to 
correspond, as shown. From the 
points in lip drop lines at right 
angles to b d, cutting the same. 
From the points in m g x carry 
lines indefinitely to the left parallel 
to the center line g f, and intersect 
them by lines of corresponding 
number erected vertically from the 
points in b d. A line traced through 
the points of intersection will give 
a correct jDlan view of the opening 
in the round end. To avoid con- 
fusion of lines the intersections 
from the points between b and h or 
the upper half of the opening are 
shown only in the near or lower 
half of the plan from t to u, while the points belong- 
ing to the lower half (/* to d) are shown only in the 
further half of the plan from p to q. 

From each of the points in p q of the plan draw 
lines to 5, which is the projection of e of the side view 
or apex of the cone in the lower half, and from the 
points in t u draw lines to o, the apex of the cone of 
the upper half of the article. These lines represent 
only the horizontal distances from s and o to the points 
in the opening t u q p of the plan or B Gr D of the front 



338 



The New Metal Worker Pattern Book. 



view. To ascertain the real distances between these 
points it will be necessary to first ascertain their ver- 
tical hights from an assumed horizontal plane and then 
to construct from these measurements a series of right 
angled triangles whose hypothenuses will give the de- 
sired distances. 

From the points in b h of the side view drop lines 
vertically, cutting a horizontal line drawn from a, as 
shown between v and j ; and from the points in h d drop 
lines to to z drawn horizontally from e. To construct 
the triangles required in the top part, first draw the 
right angle R O K, as shown in Fig. 584, and from 
on E set off the length of lines in b hjv of side view, 
as indicated by the small figures. From on K 
set off the length of lines in o t u of plan of top, also 
as indicated by the small figures. Connect the points 
in R with those of similar number in K, as shown. 
To obtain the triangles required for the bottom part, 
proceed in a similar manner. Draw the right angle 
W S L, as shown in Fig. 585. From SonSW set off 
the length of lines in h d z to of side view, as indicated 
by the small figures. From S on S L set off the length 
of lines in s p q oi plan of top, also as indicated by the 
small figures. Connect the points in S W with those 
of similar number in S L, as shown. 

For the pattern proceed as shown in Fig. 586. 
Draw the line 0', in length equal to A B of front 
view or o k of plan. Bisect 0' in C, and erect the 



D of pattern as center, and b 2 in b I of profile as radius / 
thus establishing point 2 of pattern. Proceed in this 




Fig. 584. — Diagram of Triangles in Top Half. 

manner, using the length of lines in R K for dis- 
tances from of pattern, and the stretchout between 
points in b I of profile of side view for +he distance be- 




777T5' 

Fig. 585. — Diagram of Triangles in Bottom Half. 

tween points in D G of pattern ; then draw D G and 
G: 0. With point G of pattern as center, and 5 5' in 




Fig. 5S6. — Pattern for Transition Piece Shoivn in Fig. 58S. 



perpendicular C D, in length equal to a & of side view, 
and draw D, D 0'. These lines are equal in length 
to R K of first diagram of triangles. With of lDat- 
tern as center, and 2 2' in R K as radius, describe a 
small arc, 2, which intersect with one struck from point 



W S L of triangles as radius, strike a small arc, E, 
which intersect with one struck from point of pat- 
tern as center and a e of side view as radius, thus estab- 
lishing point E of pattern. Draw Gr E and E 0. With 
point E of pattern as center, and 6 6' of triangle in 



Pattern Prohl 



ems. 



339 



W S L as radius, strike a small arc, 6, which intersect 
with one struck from point G of pattern as center, and 
I 6 of profile as radius, thus establishing point 6 of pat- 
tern. Proceed in this manner, using the length of 
lines in W S L as distance from E of pattern, and the 
stretchout between points in I d of profile of side view 
for the distance between points in G H of pattern, and 
draw G H and H E. With point H of pattern as cen- 
ter, and e d of side view as radius, strike a small arc, 
C, which intersect with one struck from point E of 
pattern as center, and ofol plan, or C of pattern, as 



radius, thus establishing point C of pattern ; then draw 
H C and C E. From E and C erect the perpendicu- 
lars E E and C F, in length equal to c e of side view, 
and draw F E. AVith of pattern as center, and a c 
of the side view as radius, strike a small arc, which 
intersect with one struck from E of pattern as center 
and e c of the side view as radius, thus establishing 

O 

point K of pattern, and draw K and K E. Then 
DGHFEEKC represents the half pattern of article. 
The other half can be obtained in the same manner or 
by duplication, as may be found convenient. 



PROBLEM 183. 

The Pattern for a Flaring Article, Round at Top and Bottom, the Top Being: Placed to One Side of 

the Center, as Seen in Plan. 



In Fig. 587, the elevation of the article is shown 
by A B D C, below which is drawn the plan of the 

A C 






10' 



^9 



5 6 

Fit,. 687. — Plan and Elevation of an Irregular Flaring Article, with 
Lines of Triangulation Shoivn in the Plan. 

dame, corresponding parts in each being connected by 
the vertical dotted lines. There are two methods of 



triangulation available in the solution of this problem 
only one of which is given in this connection. 

Divide one-half of the circle representing the base 
of the article into any convenient number of spaces, as 
indicated by the small figures, 1, 2, 3, etc. In like 
manner divide the inner circle, which represents the 
top, into the same number of spaces, as indicated by 
1', 2', 3', etc. Between the points of like numbers in 
these two circles, as for example between 2 and 2', 3 and 
3', etc., draw lines, as shown; also connect the points in 
the inner circle with points in the outer circle of the 
next higher number, as indicated by the dotted lines. 
Thus, connect 1' with 2, and 2' with 3, and so on, as 
shown. These lines just drawn across the plan are 
the bases of a number of right angled triangles whose 
altitudes are equal to the hight of the article, and 
whose hypothenuses, when drawn, will give the cor- 
rect distances across the pattern, or envelope of the 
article, between the points in the top and those in the 
bottom in the direction indicated by the lines of the 
plan. The triangles having the solid lines of the plan 
as their bases are shown in Fig. 5SS, while those con- 
structed iipon the dotted lines are shown in Fig. 589, 
and are obtained in the following manner: 

At any convenient point erect a jDerpendicular, 
E F, Fig. 58S, which in length make equal to the 
straight hight of the article, as shown in the elevation. 
From F at right angles set off a base line of indefinite 
length. On this line, measuring from F, set off lengths 
equal to the several solid lines in the plan. For 
example, make the space F 10 equal to the length 10 



310 



Tlie Neio Metal Worker Pattern Booh. 



10' in the plan, and the space F 9 equal 9 9' in the 
plan, and so on, until F 1 is set off equal to 1 1' in 
the plan. From the points thus established in the base 
line draw lines to the point E. The triangles thus 
constructed will represent sections through the arti- 
cle on the solid lines in the plan. In other words, 
the several hypothenuses of the triangles shown in 
Fi"-. 588 are equal in length to lines measured at cor- 
responding points on the surface of the completed ar- 
ticle. 

In like manner construct the triangles shown in 
Fi°\ 589, representing measurements taken on the 
dotted lines shown in the plan. Draw the perpen- 
dicular K G, equal in length to the straight night of 
the article. From K lay off a horizontal base line in- 
definite in length, drawing it at right angles to K G. 
From Iv set off lengths equal to the dotted lines in the 
plan that is, making the distance K 10 equal to the 




Fig. 58S.— Diagram of Triangles 
Based upon the Solid Lines of 
the Plan in Fig. 587. 



A?/ 

'/'/ ' I III 
/ /' / / I I 

) / \ ' \ ' \ ' / < / fa 

2 2 4 S 6 7 8 <S 



Fig. 589.— D agram of Triangles 
Based upon the Dotted Lines of 
the Plan in Fig. 587. 



distance 9' 10 in the plan, and K 9 equal to the dis- 
tance 8' 9 in the plan, and so on until K 2 is 
made equal to the distance 1' 2 in the plan. From 
the points thus established in the base line draw 
lines to the point G. Then the hypothenuses of the 
triangles thus constructed will equal measurements 
along the surface of the completed article at points 
corresponding to the dotted lines in the plan. With 
distances thus established upon the surface of the ar- 
ticle, and with the stretchout of the required pattern 
determined at both top and bottom, it is easy to lay 
out the pattern upon the general plan of constructing 
a triangle when the three sides are given. 

The development of the 'pattern can be begun at 
either end according to convenience, and the operation 
is conducted as follows : Assume any line, as 1 1' of 
Fig. 590, which make equal in length to A B of the 
elevation, or, what is the same thing, equal to E 1 of 



Fig. 588, which is one side of the first triangle. The 
other two sides are respectively the distance 1 2 of 
the plan and the hypothenuse of the triangle shown in 
Fig. 589 corresponding to the line 2 1' of the plan. 
Accordingly, take the distance 1 2 of the plan in the 
dividers, and from 1 as center describe a short arc. 
Then, taking the distance G 2 of Fig. 5S9 in the di- 
viders, and with 1' as a center, intersect the arc already 
struck, thus establishing the point 2 of the pattern. 




Fig. 590.— The Pattern of Article Shown in Fig. 587. 

The elements of the second triangle are the distance 
1' 2' on the inner circle of the plan, the hypothenuse 
of the triangle in Fig. 588 corresponding to the line 
2 2' in the plan and the side just established in the 
pattern 2 V. From 1' as center, with V 2' of the 
plan as radius, describe a short arc. Then from 2 as 
center, with E 2 of Fig. 588 as radius, describe a 
second arc, intersecting the one already made, thus 
establishing the point 2'. Proceed in this manner 
until triangles have been described adjacent to each 



Pattern Problems. 



341 



other, corresponding to the divisions first established 
in the plan. Then lines traced through the points 
thus established, as shown from 1 to 10 and from 1' to 
10', will constitute the pattern of half the article. 
The other half may be obtained by any convenient 
means of duplication and may be added on to either 
end of the half already obtained, according as it is 
desired to make the joint at the widest or narrowest 
part of the pattern. 




Fig. 591.— Model of One-Half the Article Shown in Fig. 5S7, Illus- 
trating the Construction and Use of the Triangles. 



In Fig. 591 is shown a model which may be con- 
structed of thin metal and wires, or of cardboard and 
threads, according to convenience, which will assist 
the student in forming a correct idea of the relation- 
ship existing between the various lines drawn upon 
the plan and the lines of which the pattern is con- 



structed The top and bottom of the model are dupli- 
cates of the inner and outer circles of the plan. The 
piece forming the bottom should have the solid lines 
and the inner circle of the plan drawn upon it as a 
means of placing in position the several triangular 
pieces shown, which are duplicates of the several 
triangles shown in Fig. 588. These triangular pieces 
having been placed in position according to their 
numbers, and fastened at the top and bottom, their 
outer edges, or hypothenuses, will then represent the 
solid lines drawn across the pattern, and will bear the 
same relation or angle to the edges of the top and 
bottom pieces of the model that the solid lines of the 
pattern bear to the top and bottom outlines of the pat- 
tern. Finally, threads or wires having been attached, 
as shown, will represent the dotted lines drawn across 
the pattern, and will bear the same angle to the edges 
of the solid triangles, as measured upon the model, 
that the dotted lines of the pattern bear to the solid, 
lines, as measured upon the pattern. 

Since the top and the bottom of the shape here 
shown are both round and horizontal, the figure be- 
comes that of the frustum of a scalene cone; and 
therefore, its sides, if continued upward, would ter- 
minate in an apex which can be made the common 
apex of a number of triangles whose bases are the 
spaces upon the outer line of the plan. This method 
of solving the problem as applied to a full scalene 
cone is given in Problem 163, which see. 



PROBLEM 184. 



The Pattern for a Flaring Article, Round at Top and Bottom one Side Being: Vertical. 



In Fig. 592, ABCD shows the elevation of the 
article, below which E F G H shows the plan at the 
bottom and E J K L the plan of its top, both circles 
being tangent at the point E. 

Divide the circle representing the plan of the top 
into any convenient number of equal spaces, as repre- 
sented by the small figures between K L E in the dia- 
gram. In the illustration only one-half of the plan has 
been divided, which is sufficient for the purpose. Next 
divide a like portion of the plan of the base into the 
same number of equal parts, as shown by the figures 



between E H G. Connect these two sets-, of' points^ 
first by lines drawn between like numbers, as.1 and 1', 
2 and 2', 3 and 3', etc. In a like manner- connect 1 
of the inner circle with 2' of the base, 2 with 3', 3 with 
4', etc., all as shown by the dotted lines in, the plan. 
These lines just drawn are the bases of a number of 
right-angled triangles, whose altitudes are equal to the- 
vertical hight of the article, and whose hypothenuses, 
when obtained, will give the correct measurements, 
across the pattern between the numbered, points. 

For a diagram of triangles representing the solid 



342 



Tfie New Metal Worker Pattern Book. 



lines in plan erect the vertical line P S in Fig. 593, 
equal to A B of elevation. Then at right angles from 
S lay off a base line, upon which set off distances, 
measuring from S, ecpial to the lengths of the several 
solid lines drawn across the plan in Fig. 592. Thus 
make S R ecpial to K G (1 1') and S 2 to 2 2', and so 



ELEVATION 




Fig. 592. — Plan and Elevation of Flaring Article, Showing Method of 
Triangulation. 



on. From the points thus established in the base draw 
lines to the apex. Then the hypothenuses of the tri- 
angles will be equal to measurements on the surface of 
the finished article on lines drawn from the points in the 
base to corresponding points in the top. In the same 



way construct the diagram representing the triangles 
based on the dotted lines in plan, as shown in Fig. 594. 
Set off T V equal to the straight hight of the article. 
From V draw the horizontal line V U, upon which, 




Fig. 593. — Diagram of Triangles Based upon Solid Lines in the Plan 
in Fig. 592. 



measuring from V, set off distances equal to the length 
of the dotted lines across the plan. Thus make V 2 
equal to 1 2', V 3 equal to 2 3', etc. From the points 
thus established in V U draw lines to the apex T. 
These lines will be equal to measurements upon the 



111 ^\V» 

I \\ \ v \\V\ 

\\\ \\\\\^ 

I' \ x x s \\\ 
n \ \ \\ \\ 



H 



\ \ \ \ 0\ 

I \ \ \ N \\\ 



J U I 1 1 \ \ \ 



Fig. 594. — Diagram of Triangles Based Upon Dotted Lines in the 
Plan in Fig. 592. 



surface of the finished article between the points con- 
nected by the dotted lines in the plan. 

Having obtained the correct dimensions of all the 



Pattern Problems. 



343 



triangles assumed at the beginning of the work they 
may now be constructed consecutively, thus developing 




Fig. 595.— The Pattern of Flaring Article Shown in Fig. 592. 

the pattern in the following manner : Assume any 
straight line, as D C of Fig. 595, which make equal to 



D C of the elevation, or, what is the sam t thing, PKo 1 

the diagram of triangles, Fig. 593. From C as a cen- 
ter, with a radius equal to 1' 2' of the 
plan, strike a small arc, which inter- 
sect with another small arc struck 
from D as center, with a radius equal 
to T 2 of Fig. 594, thus establish- 
ing the point 2' of the pattern. 
From D as center, with a radius 
equal to 1 2 of the plan of the top, 
Fig. 592, strike a small arc and inter- 
sect it with another struck from 2' 
of the pattern as center, and P 2 of 
Fig. 593 as a radius, thus establish- 
ing the point 2 in the top of the 
pattern. Proceed in this manner, 
using the hypothenuses of the tri- 
angles in Fig. 594 with the spaces 
in the outer curve of the plan, Fig. 
592, to establish the points in the 
bottom curve of the pattern ; and 
the hypothenuses of the triangles in 
Fig. 593 with the spaces in the inner 
curve of the plan to establish the 
points in the top curve of the pat- 
tern. Lines traced through the 
points of intersection, as shown from 
C to B and from D to A, will, with 
D C and A B, constitute the pattern 
for the half of the article shown by 
E II G K of the plan. The other 
half may be added, as shown, by 
any convenient means of duplica- 
tion. 

Since the top and the bottom 
of this figure are both round and 
horizontal, it becomes the frustum 
of a scalene cone, which permits of 
its being treated by a different, and 
perhaps simpler, method of triangu- 

lation, all of which is given in Problem 1 67, to which 

the reader is referred. 



PROBLEM 185. 

The Pattern of an Article having an Elliptical Base and a Round Top. 



Fig. 596 shows the plan and elevation of the ar- 
ticle for which the pattern is required. Divide one- 
quarter part of the plan of the base E G into any 



convenient number of equal spaces, and divide a cor- 
responding part of the plan of the top L K into the 
same number of spaces, numbering the points of divi- 



344 



The New Metal Worker Pattern Book. 



sion the same in both, as indicated by the small figures 
1, 2, 3 and l 1 , 2 1 , 3 1 , etc. The article here shown pos- 
sesses some of the general features of the cone in that 
it is tapering in its sides, but inspection will show that 
the slant or taper of its sides varies in different parts 
of its circumference, or in other words, that different 




Fig. 596. — Elevation and Plan of Flaring Article with Elliptical 
Base and Round Top. 

lines drawn through like numbers in the base and top 
would, if extended upward, meet the axis at different 
hights, hence some means must be devised for meas- 
uring the real distances between the points in the base 
and the points in the top, which may be accomplished 
in the following manner : First connect all points in 
the base in plan with points of the same number in 



M 

ft," 

//hi 
/, a ii 

1 1 a 




13 3 4 6 6 7 







V 



Fig. 597. — Diagram of Triangles 
Based upon Solid Lines of the 
Plan in Fig. 596. 



2 3 1 607 

Fig. 598. — Diagram of Triangles 
Based upon Dotted Lines of 
the Plan in Fig. 596. 



the top by means of a solid line, as shown upon the 
plan by lines 1, l 1 , 2, 2 1 , etc. Also draw the inter- 
mediate dotted lines connecting alternate points, as 
shown in the engraving by 2 l 1 , 3 2 1 4 3 1 , etc., thus 
dividing the entire surface of the article into triangles. 
Construct a diagram, as shown by A 1 N 1 C, Fig. 
597, in which the actual distance between correspond- 
ing points in base and top shall be shown. Make 
C N 1 equalto the straight hight of the article, C 1ST of 



the elevation. At right angles to it set off N 1 A 1 , in 
length equal to the distance I 1 1 in plan. From N' 
set off also on N 1 A 1 spaces corresponding to 2 1 2, 
3' 3, 4 1 4, etc., of the plan, and from each of these 
points draw a line to C, as shown. Then the lines 
converging at C 1 represent the distances which would 
be obtained by measurements made at corresponding 
points upon the article itself. Construct a like dia- 
gram of the distances represented in the dotted lines 
in the plan, as shown by C 2 W 0, Fig. 598. Make 
C 2 N 2 equal to C N" of the elevation, and from N 2 set 
off at right angles the line N 2 0. Upon this line make 
the spaces ISP 2, 1ST 2 3, N 2 4, etc., equal to the length 



of the dotted lines l 1 



V ?. 



3 1 4, etc., and from the 




Fig. 599.— The Pattern of Article Shown in Fig. 596. 

points thus obtained in N 2 draw lines to C 2 . Then 
these converging lines represent the same distances as 
would be obtained if measurements were made between 
corresponding points upon the completed article. 

For the pattern, commence by drawing any line, as 
P X in Fig. 599, on which set off a distance equal to 
C 1 of the first diagram, as shown by 1 l 1 . Then, 
with the distance from 1 to 2 of the plan for radius, and 

I in pattern as center, describe an arc, which intersect 
by another arc struck from I 1 of the jDattern as center, 
and C 2 2 of the second diagram as radius, thus estab- 
lishing the point marked 2 in the pattern. Next, with 

I I 2 1 of the plan as radius, and from l 1 of the pattern 
as center, describe an arc, which intersect by another 



Pattern Problems. 



345 



arc drawn from 2 as center, and with C 2 of the first 
diagram as radius, thus locating the point 2 1 of the pat- 
tern. Continue in this manner, locating each of the 
several points shown from X to Y and from P to R of 
the pattern, through the several intersections, tracing 



the lines of the pattern, as shown. Then XYRP 
will be one-quarter of the required pattern. Repeat 
this piece three times additional, as shown by W X P 
T, V W T IT and Y Z S R. reversing each alternate 
piece, thus completing the pattern. 



PROBLEM 186. 

Pattern for an Irregular Flaring; Article which is Elliptical at the Base and Round at the Top, the Top 
being - so Situated as to be Tangfent to One End of the Base when Viewed in Plan. 



In Fig. 600, let D G F E be the side elevation of 
the article and KNM one-half of the plan of the base. 
The half plan of the top is shown by K W L, the base 
and top being tangent in plan at the point K. The 
conditions and method of procedure in this problem 
do not differ materially from those of Problem 184. 



Next construct the diagrams of triangles, as shown 
in Fig. 001 at. the right of the elevation, making A U 
in hight equal to D E of the elevation, and lay off U T 
at right angles to it. Let A represent all points in the 
circle representing the plan of the top of the article. 
Lay off from U upon U T the distance from each of 




\\\\ 

i \ \ \ \ 

I \ \ \ 

1 \ \ \ \ 



Fig. 601. — Diagrams of Triangles Obtained 
from Fig. 600. 



Fig. 600.— Elevation and Plan of an Irregular Flaring Article 
with Elliptical Base and Round Top. 



The surface of this article may be divided into 
measurable triangles in the following manner : Divide 
the plans of top and base into the same number of 
equal parts, as shown by 1, 2, 3, etc., in the base and 
l l , 2 1 , 3 1 , etc., in the top, and connect similar points in 
the two by solid lines, as shown by 6 6 1 , 5 5 1 , etc. Also 
connect each point in the plan of the top with the next 
lower number in the plan of the base, as shown by the 
dotted lines in the engraving, as 6 7 1 , 5 6 1 , etc. 



the several points in the circle to the corresponding 
point in the ellipse. Thus make U 7 equal to 7 1 7 of 
the plan, IT 6 equal to 1 0, etc. , and draw the radial lines 
A 2, A 3, etc. In like manner construct a correspond- 
ing diagram, as shown by C B V, using for the spaces 
in B V the lengths of the dotted lines between the circle 
and the ellipse in the plan, and draw C 2, C 3, C 4, etc. 
By means of these two sets of lines, converging at A 
and C respectively, and the stretchouts of the two 



346 



Ttie New Metal Worker Pattern Book. 



carves of the plan, the actual dimensions of the tri- 
angles into which the surface of the article has been 
divided can be accurately measured. 

These are to be used in describing the pattern as 
follows : At any convenient place draw the straight 
line P R in Fig. 602, in length equal to G F of the 
elevation, or, what is the same, equal to A 7 of the 
first diagram. As but half of the plan of the article 
is shown, the pattern will also appear as one-half of 
the whole shape, and therefore P R will form its cen- 
tral line. From P as center, with radius C 6 of the 
second diagram, describe an arc, which intersect by a 
second arc struck from R as center, with radius 7 6 of 
plan, thus establishing the point 6 of the pattern. 
Then with radius A 6 of the first diagram, from 6 of 
the pattern as center, describe an arc, which cut with 
another arc struck from 7 1 of the pattern as center, and 
7 1 6 1 of the plan as radius, thus locating the point 6' of 
the pattern. Continue this process, locating in turn 
5, 5', 4, 4 1 , etc., until points corresponding to all the 
points laid off in the plan are established. Lines traced 




Fig. 602. — Pattern for Article shown in Fig. 600. 

through the points 7, 6, 5, etc., and 7 1 , 6 1 , 5 1 , etc., 
will, with P R and O S, form the patter^ of one-half 
the article. 



PROBLEM 187. 

Pattern lor a Flaring Article or Transition Piece Round at the Top and Oblong at the Bottom, the 

Two Ends Being: Concentric in Plan. 



ELEVATION 
C 



END ELEVATION 




Fig. 60S.' 



Q \^_ ^_^ 

P 

PLAN 
-Plan and Elevations of Flaring Article Round at the Top 
and Oblong at the Bottom. 



In Fig. 603 are shown the side 
and end elevations and the plan of an 
article which might form a transi- 
tion between an oblong pipe below 
and a round pipe above. According 
to the conditions, as given in the en- 
graving, the problem is capable of 
two solutions. Since the upper and 
lower bases are composed, either 
wholly or in part, of semicircles 
lying in parallel planes, those por- 
tions of the pattern of the article lying between the 
semicircles, asPONOP, must necessarily form parts 
of the. envelope of a scalene cone. Those portions of 
the pattern may therefore be obtained, if desirable, 
by the method employed in Problems 168 and 169. 

The other solution, which is perhaps the more 
simple, is given in Figs. 604 to 606. An inspection 
of the plan will show that the article consists of four 
like quarters, therefore in Fig. 604 is shown an en- 
larged plan and elevation of one-quarter of the article. 
Divide P T and O N of the plan each into the 



Pattern Problems, 



347 



same number of equal parts, and connect the points 
in P T with those in N, as indicated by the solid 
lines. Also connect points in the top with those in 



of dotted lines in plan. Thus make "W 7 equal to T 7 
and W 8 equal to 2 8, and so on. From the points 
thus established in the base draw lines to V. The 





K H 6 7 89io L w 

Fig. 605.— Diagram of Triangles. 



98 7)( 
10 



Fig. 604.— Quarter Plan and Elevation Enlarged, Showing Method 
of Triangulation. 

the bottom, as shown by the dotted lines. These lines 
represent the bases of right angled triangles, the alti- 
tude of which will be equal to the straight hight of the 
article. For a diagram of the triangles representing the 
solid lines of the plan, draw any vertical line, as J K in 
Fig. 605, which make equal in hight to the hight of the 
article, as shown by the dotted line D F. From K, at 
right angles to J K, draw K L, upon which set off dis- 
tances, measuring from K, equal to the lengths of the 
solid lines drawn across the plan. Thus make K 6 
equal to T 1ST and K 7 equal to 2 7 of plan, and so on. 
Also set off from K the distance H P of plan, as 
shown by K H. From the points thus established in 
K L draw lines to J. The hvpothenuses thus obtained 
will give the distances across the finished article, as in- 
dicated by the solid lines of the plan. 

The next step will be to construct a diagram of 
triangles that will give the distances between points in 
the base and top, as indicated by the dotted lines in 
plan. This diagram is constructed in a similar man- 
ner, as shown at the right in Fig. 605. Draw the right 
angle VWX, making V W equal to the straight hight 
of the article, and from W set off on W X the lengths 



hypothenuses of the triangles thus obtained win give 
the distance from points in the base to jaoints in the 
top, as indicated by the dotted lines in plau. 

To lay out the pattern first draw any line, as T N 
of pattern, in length equal to J 6 of first diagram of 
triangles, or, which is the same thing, D E of 
elevation. From N of pattern strike a short arc 



2t ■ 



\\ \ 



10 


Fig, 60S.— Quarter Pattern of Article Shown in Fig. 60S. 

with a radius equal to N 7 of the plan, as shown. 
From T of pattern as center, with radius equal to V 7 
of the second set of triangles, intersect this arc, thus 
establishing the point 7 of pattern. From T, with 
radius equal to T 2 of the plan, strike a small arc, as 
shown, and intersect it with another from point 7 of 
pattern as center, with J 7 of the diagram of triangles 
as radius, thus establishing the point 2 in pattern. 



348 



Hie New Metal Worker Pattern Book. 



Proceed in this manner, using alternately the hypoth- 
enuses of the triangles in V W X of Fig. 605, the 
spaces in plan of base N, the hypothenuses of the 
triangles in J K L, Fig. 605, and. the spaces in the plan 
of top, P T, ill the order named, and as above ex- 
plained. The resulting points, as indicated by the 
small figures in the pattern, will be points through 



which the pattern line will pass. For the pattern of 
triangle P H of pattern, with of pattern as center, 
and H of plan as radius, strike a small arc in the 
direction of H. With P of pattern as center, and J H 
of the diagram of triangles as radius, describe another 
small arc intersecting the one just struck. Draw H 
and H P, thus completing the quarter pattern. 



PROBLEM 188. 
Pattern for a Transition Piece Round at the Top and Oblong at the Bottom. Two Cases. 



In Fig. 607 is shown the plan and elevations 
of a transition piece, constituting the first case, such 
as is frequently required in furnace work when it 
is necessary to connect a round pipe with another 
pipe of equal area but flattened into an oblong 
shape. 

la Fig. 611 are shown the plan and elevations of a 



SIDE VIEW. 




98765432! 



101112 13 



first case, shown in Fig. 607. The principle involved 
in developing the patterns of the two shapes is exactly 
the same, consequently the following demonstration 
will apply equally well to either Fig. 607 or 611, in 
each of which corresponding points are lettered the 
same. (It will be noticed that separate diagrams of 
triangles and a separate pattern corresponding to Fig. 




Fig. 608.— Diagrams of Triangles Based upon the Solid Lines of \_ Q 

the Plan, Fig. 607. Fig _ 607.— Plan and Elevation of Transition Piece— First Case. 



transition piece of the second case, answering the same 
general description as that given above, but differing 
only in the fact that the circle representing the top in 
the plan view does not touch the side of the line rep- 
resenting the bottom of the article. In other words, 
the side F H is slanting instead of vertical, as in the 



611 have not been given. While in reality they would 
differ somewhat from those shown in Figs. 60S to 610, 
they would have the same general appearance, and in 
method of construction would be exactly the same, and 
therefore have not been considered necessary to the 
study of the problem.) 



Pattern Problems. 



349 



N P Q represents a plan of the shape described, 
above which A B D C shows an elevation of the front 
of the same, or as seen when looking toward Q, while 
to the left is shown a view obtained by looking to- 
ward the side N, in which E Gr corresponds to P X of 
the plan and F H to Q 1 Q, 

Divide one-half of the plan of the top or round 
portion of the article into any convenient number of 
equal spaces, in this case 13. Since by the conditions 
of the problem one-half of the round end corresponds 
to the semicircular end of the oblong part, divide the 
semicircle J 1ST L into the same number of equal parts. 









\\\\v 



\\ \ \\\ 



6 5 4 3 2 1 



7 8 9 10 11 12 



Fig. 609.- 



-Diagrams of Triangles Based upon the Dotted Lines 
of the Plan, Fig. 607. 



Then connect points in the two lines of the plan of th 
same numerals. For example, 1 with 1, 2 with 2, 
with 3, etc. In like manner connect the 
points in the end of the oblong portion 
with points of the next higher num- 
ber in the round end, as shown, as, 
for example, 1 with 2, 2 with 3, 3 with 
4, etc. Upon all of these lines drawn 
in the plan it will be necessary to 
construct sections or triangles in which 
these lines form the bases and in 
which the vertical hight of the article 
W V is the altitude. The various 
hypothenuses thus obtained will then 
represent the true distances across the 
finished article upon the lines indicated 
in the plan. The triangles correspond- 
ing to the solid lines of the plan are shown in Fig. 60S, 
while those corresponding to the dotted lines of the plan 
are shown in Fig. 609. In order to avoid confusion each 
of these sets has been divided into two groups, as shown, 
and are constructed as follows : Lay off at any con- 
venient place the line A B (Fig. 608), equal to V W 
of Fig. 607. From the point B, and at right angles to 
A B, draw the line B C and upon it set off the lengths 



of the several solid lines connecting the two outlines 
in the plan. Thus make B 1 equal to the distance 1 1 
or J P of the plan. B 2 is equal to the distance 2 2 
of the plan, and B 3 is equal to the distance 3 3 of the 
plan, etc. As already explained, D E is a duplicate 
of A B, and E F is drawn at right angles. On E F 
the spaces E 10, E 11, E 12 and E 13 are set off, 
being equal respectively to 10 10, 11 11, etc., of the 
plan. From the points thus established in the base 
lines B C and E F draw lines to the apices A and D, 
thus completing the triangles. Then the hypothenuses 
A 1, A 2, A 3, etc., I) 13, D 12, etc., correspond to 
tin.' width of the pattern measured between points in- 
dicated by like figures in the plan. 

In the same general manner construct the triangles 
shown in Fig. 609, which correspond to sections on 
the dotted lines across the plan. Gr H and L M of Fig. 
609 correspond to the hight V W of the elevation. 
H K and M N" are drawn at right angles to the perpen- 
diculars, and on these base lines spaces are set off, 
measuring from H and M respectively, corresponding 
to the length of the clotted lines across the plan. Thus 
H 1 corresponds to 1 2 of the plan, and M 12 corres- 
ponds to 12 13 of the plan. From the points thus estab- 
lished in the base line lines are drawn to the apices Gr 
and L, thus completing the triangles. These hypothe- 
nuses are equal to the width of the pattern measured 




1 1 1 2" 13 

Fig. 610. — Pattern of Transition Piece Shown in Fig. 607. 



between points connected by the dotted lines in the 
plan. By the conditions of the problem, inasmuch as 
there are straight portions in the oblong end, there will 
be portions of the pattern that will correspond to tri- 
angles the bases of which are equivalent to the length 
of the straight portion in the plan and the hights of 
which are equal respectively to the distances E G and 
F H of the side view. 



350 



Tlie New Metal Worker Pattern Book. 



SIDE VIEW 



Therefore, to describe the pattern proceed as fol- 
lows : At any convenient place, as shown by A B in 
Fig. 610, draw a line equal to the width of the pattern 
at a point corresponding to Q 1 Q in the plan. This 
would be the same 
as F H of Fig. 607 E_ 
or 611. To com- 
plete the triangu- 
lar portion referred 
to set off from B 
the distance B C 
equivalent to Q L 
of Fig. 607, thus 
obtaining the point 
C. The dotted line 
A C in the pat- 
tern is drawn to show the portion obtained by this 
means. From C as center, with the space 13 12 of the 
plan of the oblong end as radius, describe a small arc, as 
shown to the left. Then from A as center, with radius 
L 12 of Fig. 609, corresponding to the width of the pat- 
tern measured on the dotted line 13 12 of the plan, de- 
rscribe another arc intersecting the one just drawn, thus 
establishing the point 12 in the lower edge of the pat- 
tern. From 12 as center, with D 12 of Fig. 608 as 
radius, being the width of the pattern on the line 1212 
of the plan, describe a short arc, as shown at 12 in the 
upper line in the pattern. Intersect this with another 
arc drawn from A as center, with 13 12 of the plan of 
the round end as radius, thus locating the point 12 in 
the upper line of the plan. Proceed in this manner, 
using in the order described the stretchout of the semi- 
circular end of the oblong section, the hypothenuses of 
the triangles corresponding to the dotted lines in the 
plan, the hypothenuses of the triangles corresponding to 
the solid lines in the plan and the stretchout of the 
circular end, reaching finally the points D and E of the 
pattern, representing one side of the remaining triangu- 



lar section to be added. From E as center, with K J 
of the plan as radius, describe an arc. From D as 
center, with radius equal to D E of the pattern, strike 
a second arc intersecting the one just drawn, thus 




Fig. 



13 l Q M 

611.— Plan and Elevation of Transition Piece — Second Case. 



locating the point F. Connect F and E and also F and 
D, thus completing the pattern of the part correspond- 
ing to K J P of the plan. The dotted line D G drawn 
across the pattern corresponds to the line X P of the 
plan, and DA B G will be one-half of the finished 
pattern. 



PROBLEM 189. 

Pattern for an Offset Between Two Pipes, Oblong; in Section, whose Long" Diameters Lie at Riglit 

Angles to Each Other. 



In Fig. 612 are shown the plan and elevations of 
an offset or transition piece to form a connection be- 
tween two pipes .of oblong profile which will be spoken 
of in the demonstration as the upper and the loiuer 
pipes. MNOPQL of the plan is the section or 



profile of the upper pipe which begins at the line A B 
of the elevation and extends upward, while F Gr H I 
J K of the plan is the section of lower pipe which be- 
gins at the line D of the elevation and extends 
downward, A B C D being one view of the offset. At 



Pattern Problems. 



35L 



the right the same plan is shown turned one-quarter 
way around, from which, and the front elevation, are 
projected a side elevation of the offset. Corresponding 
points in the two plans are indicated by the same let- 
ters, capitals being used in the one and italics in the 
other. 

An inspection of the plan will show that the long 
diameter O L of the upper pipe cuts the profile of the 
lower pipe nearer one end than the other, from which 



the offset. Lines from G H of the bottom to N - of 
the top would form another corner, lines from I J up 
to P a third corner and lines from J K up to Q L 
the fourth. Lying between these corner pieces are 
the two triangular end pieces K L F and I II and 
the side pieces M N G and Q P J. For convenience 
in describing the pattern the joint will be assumed 
through one of the ends at the line E L. 

Preparatory to obtaining the pattern, first divide 




o 

PLAN 



Fig. 612.— Plan and Elevations of an Offset betiveen Two Pipes of 
Oblong Profile. 



it must be concluded that the pattern cannot be com- 
posed of symmetrical halves or quarters and that there- 
fore the entire pattern must be developed at one oper- 
ation. 

As the sections of both pipes may be said to con- 
sist of four quarter circles joined to the straight side, 
the pattern will consist principally of four rounded 
corners joining the quarter circles which occupy the 
same relative position in the two pipes. Thus lines 
joining the quarter circles F G of the bottom and L M 
of the top would form one corner of the envelope of 



each of the four quarter circles of the lower pipe into 
the same number of equal parts ; also divide the pro- 
file of the upper pipe in the same manner. To avoid 
confusion of lines two separate plans aro shown in 
Figs. 613 and 614 for obtaining these divisions. In 
Fio-. 613 are shown the divisions of what may be called 
the back end, while Fig. 614 shows those of the front 
end. As will be seen by these plans, the points have 
been numbered alternately in the bottom and the top. 
Solid lines are first drawn, as shown, connecting points 
12 3 4, 5 6, etc. , after which the four-sided figures thus 



352 



The New Metal Worker Pattern Booh. 



produced are divided diagonally by the dotted lines 
1 4, 3 6, 5 8, etc. 

The next operation is the construction of a series 



between F and G in the lower pipe to points between 
M and L of upper pipe, as indicated by the solid lines 
in plan. For the diagram of triangles representing the 



PLAN 





Fig-. <iU. 



Duplicates of Plan in Fig. 612, Showing System of Triangulation. 



of right angled triangles whose bases shall be equal to 
the several solid and dotted lines which have just 
been drawn, and whose altitudes shall equal the straight 
night of the offset. These have been arranged in four 
groups, shown in Figs. 615 to CIS, corresponding to 
the four corner pieces above mentioned. In Fig. 615, 
draw the right angle R S T, making R S equal to the 
straight hight of the article, as indicated by A Z of 
front elevation. Measuring from S, set off on S T the 
lengths of solid lines in F G M L of the plan Fig. 613, 
including L E, which will give the slant hight cor- 

Uk. 







dotted lines in FGML of the plan, draw the right 
angle R' S' T 1 , as shown at the right in Fig. 615, 
making R 1 S 1 equal to the straight hight of the article, 
as derived from A Z of front elevation. Measuring in 



r! 




\ 



% 



\\S\ 



^ 



Fig. 616.— Diagrams of Triangles in G H O N of Fig. 614. 



responding with b c of the side elevation. Connect the 
points in S T with R, as shown by the solid lines. 
These hypothenuses represent the distances from points 



lines 
The 



1367 



Fig. 615.— Diagrams of Triangles in FGML of Fig. 613. 

each instance from S 1 , set off on 
S 1 T 1 the lengths of dotted lines in 
FGML, and from the points thus 

W obtained draw lines to R 1 . The 

i 

diagrams shown in Fig. 616 are 

constructed in the same manner and 

correspond to the solid and dotted 

the corner G H N, shown in Fig. 614. 

of triangles in Fig. 617 are derived 



is Is 
I ii 



diagrams 



from the solid and clotted lines in I J P of Fig. 



Pattern Problems. 



353 



Gil, and in Fig. 61 S are sliown the diagrams of tri- 
angles derived from the solid and dotted lines in Q L 
K J of Fig. 613. 



the plan as 
Proceed in 
dicated by 







^§ 



19 21 1 66 



Fig. 617. — Diagrams of Triangles in IJPO of Fig. 614. 

To develop the pattern draw any line, as E L of 
Fig. 610, in length equal to E E of Fig. 615, which 
gives the actual distance from E in the base to L in the 
top, as also shown by b c of side elevation and indicated 
in corresponding plan by I e. With E of pattern as 
center, and E F of plan as radius, strike a small arc, 
F, which intersect with one struck from L of pattern 
as center, and E T of Fig. 615 as radius, thus estab- 
lishing point F of pattern. With point F of pattern 
as center, and E 1 1 of Fitr. 615 as radius, strike a 
small arc, 1, which intersect with one struck from 
point L of pattern as center, and L -1 of Fig. 613 as 



radius, thus establishing point 3 of pattern, 
this manner, as above described, and as in- 
the solid and dotted lines, until the points G 
and M of pattern are located. With M 
of pattern as center, and M N of the 
plan as radius, strike a small arc, N, 
which intersect with one struck from 
G of pattern as center, and U W of 
Fig. 616 as radius, thus establishing 
point N of pattern. Proceed in the 
manner indicated until the remaining 
points in the pattern are located. It 
will be observed that the letters and 
figures in pattern designate points 




^*\ 

y 

33 [ -a 

I 29 



Fig. 61S. — Diagrams of Triangles in Q L K J of Fig. 613. 



\ 







619.— Pattern of Offset Shown in Fig. 612. 



radius, thus establishing point 4 of pattern. With 
point 4 of pattern as center, and E 4 of Fig. 615 as 
radius, strike a small arc, 3, which intersect with one 
struck from point F of pattern as center, and F 3 of 



similarly indicated in Figs. 613 and 614. Lines traced 
through the points obtained as directed, and as 
shown from E to E' and L' to L, will produce the de- 
sired pattern. 



354: 



Tlte Neiv Metal Worker Pattern Book. 



PROBLEM 190. 

Pattern for an Irregular Flaring Article Whose Top is a Circle and Whose Base is a Quadrant. 



In Fig. (320 G E F shows the plan of the article 
at the base, L J K the plan at the top and ABCD 
an elevation of one side. An inspection of the plan 
•will show that the article consists of two symmetrical 
halves when divided by the line G H, and that, there- 




G F 

Fig. 620. — Plan and Elevation of an Article Whose Top is a Circle 
and Whose Base is a Quadrant. 

foie, the triangulation of one-half will answer for the 
whole. On account of the dissimilarity between the 
outlines of the top and the bottom some judgment will 
be required in adopting a good division of the surface 
into triangles. 

As the point L of the plan is the nearest point to 
the adjacent side E G, it must be chosen as the vertex 
of a triangle whose base is E G. That portion of the 
circle of the top, therefore, between L and its corre- 
sponding point K in the other half of the article must 
be considered as the base of an oblique cone whose 
apex is at G. 

It is always advisable in the division of a surface 
into triangles that the solid and dotted lines crossing 
the plan should intersect the outlines of the top and 
the bottom as nearly at right angles as possible. 



Therefore, since the remainder of the top (L to J) and 
E H of the base are the bases of a surface which must be 
so divided as to best serve the purposes of triangulation, 
it is advisable to divide L J into more spaces than 
E H, allowing the extra spaces in the top nearest the 

point L to form the bases of a 
number of converging triangles, 
as shown. Thus first divide E 
H into any suitable number of 
spaces, as shown by the small 
figures 5 to 9, then divide L 
J into a greater number of equal 
spaces than E H, as shown by 
the small figures 3 to 9. Con- 
nect points of like number in 
the two outlines by solid lines, 
commencing at H and J, as 
shown from 9 9 to 5 5, drawing 
lines also from 4 and 3 of the top to 5 (E) of the bot- 
tom. Also draw the dotted lines 5 6, 6 7, etc., and 
the solid lines from points in L N to G. These solid 
and dotted lines will then form the bases of a series of 
right angled triangles whose hypothenuses will give 
the real distances across the envelope of the finished 
article. 

These triangles are constructed, as shown in Fig. 
621 at the right of the elevation, in the following man- 
ner. Extend A B and D C of the elevation, through 
which draw any vertical line, as Q R. From Q on Q P 
set off the lengths of all the solid lines of the plan. 
Thus make Q 9 equal to 9 9 or J H of the plan, Q 8 
equal to 8 S of the plan, etc. , and from the points thus 
established draw lines to R. In like manner draw the 
vertical line T V, and from T on T S set off the 
lengths of the dotted lines of the plan, as shown by 
the small figures, and from the points thus obtained 
draw lines to V, as shown. The small figures in S T 
correspond with the figures in L J, the top line of the 
plan. 

In laying out the pattern shown in Fig. 622 the 
joint is assumed upon the line J H of the plan. The 
pattern may be best begun by first laying out one of 
the large triangles forming a side of the article, as- 
E L G or G K F of the plan, shown also by D N C of 
the elevation, Fig. 620. Draw any horizontal line, as- 
E G of Fig. 622, equal in length to E G of the plan. 



Pattern Problems. 



355 



From E as center, with radius R 3 of Fig. 621, de- 
scribe a small arc near L, which intersect with another 
arc drawn from G as center, with a radius equal to 
R 3' of Fig. 621, thus establishing the point L of the 
pattern. From G of the pattern as center, with radii 




E G 

Fig. 622.— Pattern of Article Shown in Fig. 620. 

equal to R 2 and R lof Fig. 621, describe small arcs, as 
shown between L and K of the pattern. Take between 
the points of the dividers a space equal that used in 
dividing the arc L K of the plan, and placing one foot 
of the dividers at L of the pattern step from arc to 
arc, reaching K, as shown, and through the points 
thus obtained draw L K of the pattern ; also draw 
K G. From E of the pattern as center , with radii equal 
to R i and R 5 of Fig. 621, describe small arcs to the 



left of L. With the dividers set to the space used in 
dividing the arc L J of the plan, place one foot at 
L (3) and step first to arc 4, then to arc 5, thus es- 
tablishing the points 4 and 5. 

With the last obtained point, 5, of the pattern as 
center, and a radius equal to the 
dotted line V 5 of Fig. 621, describe 
a small arc (6'), which intersect with 
another arc struck from point E of 
pattern as center, with a radius equal 
to 5 6 of the base line E H of the 
plan, thus establishing the point 6' 
of the pattern. With 6' of the pat- 
tern as center, and a radius equal to R 
5 of Fig. 621, describe a small arc 
(6), which intersect with another arc 
struck from 5 of pattern as center, 
with a radius equal to 5 6 of the 
top line L J of the plan, thus estab- 
lishing the point 6 of the pattern. 
Proceed in this manner in the con- 
struction of the remaining; triangles 
of the pattern, using alternately the 
lengths of the dotted and the solid 
hypothenuses in Fig. 621 corre- 
sponding to the dotted and the solid lines crossing the 
plan, in the order in which they occur, to determine 
the width of the j:>attern ; the spaces in E H of the 
plan to form the lower line E H of the pattern and 
the spaces in L J of the plan to form the upper line 
of the pattern, all as shown. The remaining parts 
of the pattern can be obtained by any convenient 
means of duplication, K F G being a duplicate of 
L E G and K J 1 H' F being a duplicate of L J H E. 



PROBLEM 191. 



The Patterns for a Three-Piece Elbow, the Middle Piece of which Tapers. 



In Fig. 623, let A B D F H G E C be the side 
view of a three-piece elbow, the middle piece (C D F E) 
of which is made tapering. The piece CDFE may 
also be described as an offset between two round pipes 
of different diameters. A half profile of the upper 
and smaller of the two pipes is shown by a m b, while 



g n h shows that of the larger pipe. * The straight por- 
tions ABDC and E F H G are in all respects similar 
to many pieces whose patterns have already been de- 
scribed in the first section of this chapter in Problems 
38 to 45 inclusive. It will therefore be unnecessary 
to repeat the description in this connection. 



6i> 



6 



TJie New Metal Worker Pattern Booh. 



Since an oblique section through a cylinder is an 
ellipse, an insjuection of the drawing will show that the 
sections C D and E F, the upper and lower bases of 
the middle piece, must be elliptical. The first opera- 
tion, therefore, will be to develop the ellipses, which 
may be done in the following manner : Divide the 




■ \\ i\ — i-V 



\ 

\\ 

\ 

H \ \\ 



\ 




Fig. 623 



-Elevation of a Three-Piece Elbow, the Middle Piece of 
which Tapers. 



profile a m b into any convenient number of equal 
spaces, as shown by the small figures. From the points 
thus obtained drop lines vertically to a b and continue 
them till they cut the line C D. From the intersec- 
tions on C D carry lines at right angles to the same 
indefinitely, as shown. Through these lines draw any 



line, as c d, parallel to C D. Upon each of the lines 
drawn from C D, and measuring from c d, set off the 
lengths of lines of corresponding number in the profile 
a m b, measuring from a b. A line traced through the 
points thus obtained, as shown by c k d, will be the 
required elliptical section. The section upon the line 
E F may be obtained in the same manner, all as shown 
by epf. 

In Fig. 624, C D F E is a duplicate of the middle 
piece of Fig. 623, below which is drawn its half plan 
made up of the elliptical sections just obtained, all as 
shown by corresponding letters. The piece thus be- 




Fig. 624.— The Middle Piece of Elbow in Fig. 623, with Plans of 
its Bases Arranged for Triangulation. 



comes an irregular flaring article or transition piece? 
the envelope or pattern of which may be obtained in 
exactly the same manner as described in Problems 1S4 
or 186, to which the reader is referred. The eleva- 
tion in Fig. 623 is so drawn that C D and E F are 
parallel, and C E is at right angles to both. Should 
the elevation, however, be so drawn that C D is not 
parallel with E F the conditions will then become the 
same as in Problem 193 succeeding, which see; and 
should C E be drawn otherwise than vertically, the 
plan would then resemble that shown in Prob- 
lem 194. 



PROBLEM 192. 

The Patterns for a Raking: Bracket in a Curved Pediment. 



In Fia;. 625, let C E F D be the front elevation of 

o 7 

a portion of a curved pediment whose center is at K, 
and of which E K is the center line. C A B D of the 
same elevation represents the face view of a bracket 
having vertical sides, of which E G F is the normal 



profile. Since the bracket sides are vertical and are 
necessarily at different distances from the center line, 
it will be easily seen that they are of different lengths 
or hights ; that is, the side C D, being further from 
E K than the side A B, is longer. The patterns for 



Pattern Problems 



357 



the two sides will therefore be different and the face 
piece will be really an irregular flaring piece. 

It will first be necessary to obtain the pattern or 
profiles of the two sides. To facilitate this operation 
the normal profile of the bracket E G F has been so 



line E F, as shown. Thence carry lines around the 
arch from the center K, from which the same is struck, 
cutting the sides A B and G D. Conveniently near 
the lower side of the bracket draw any vertical line, as 
E 1 B', as a base line upon which to construct the true 




RIGHT SIDE 
ELEVAT 



Fig. 625.— A Raking Bracket in a Curved Pediment, Showing the Patterns for Its Face and Sides. 



placed that its vertical line or back coincides with the 
, center line E K of the arch. Divide the face of this 
profile into any convenient number of parts, as shown 
by the small figures, and from these points carry lines 
at right angles to the back of the bracket, cutting the 



sides of the bracket. At any convenient position 
upon this line, above or below, as at Gr E 1 F 1 , draw a 
duplicate of the normal profile, so that its back or 
vertical line shall coincide with E 1 B 1 , and divide its 
face line into the same spaces as Gr F. Place the 



358 



Tlie New Metal Worker Pattern Booh. 



T-square at right angles to the line E 1 B 1 , and, bringing 
it successively against the points in the side C D, 
draw lines cutting E 1 B', continuing the same indefi- 
nitely to the left, as shown. At any convenient po- 
sition, as A 1 B 1 , on the line E 1 B 1 transfer the spaces 
from A B, as shown, and from the points thus ob- 
tained draw lines indefinitely to the left also at right 
angles to E' B'. Place the T-square parallel to E' B 1 , 
and, bringing it successively to the points in the normal 
jn'ofile G 1 F l , cut lines of corresponding number in the 
two sets of parallel lines just drawn. Lines traced 
through the points of intersection will give the re- 
quired patterns of the lower and upper sides, as shown 
respectively by C M D 1 and A 1 N B'. Some of the 
lines of projection in the pattern of the upper side 
have been omitted to avoid confusion. At the ex- 
treme left of the engraving is shown a side view of 
the bracket as seen from the right, which is made up 
of the two sides just obtained and which have been 
placed in proper relation to each other, all as shown 
by the dotted lines projected to the left from the 
points A, B, G and D of the front elevation. 

Having now obtained all that is necessary, it re- 
mains to triangulate the face of the bracket preparatory 
to developing the pattern of the same. With this in 
view first connect all points of like number in the 
upper and lower sides of the front view by solid lines, 
as shown. Also connect them in the side elevation. 
Since points of like number in A B and C D have the 
same projection from the back of the bracket, it will 
be seen that the solid lines just drawn connecting 
them represent true distances across the face of the 
bracket. The four-sided figures produced by draw- 
ing these lines must now be subdivided into triangles 
by means of clotted lines drawn diagonally through 
each. Therefore connect each point uj>on the profile 
of the lower side of the bracket with the point next 
higher in number upon the upper side, as shown in 
the side view. To determine the true length of these 
lines it will be necessary to construct a diagram of 
triangles, as shown by S V T in the upper part of the 
engraving. Draw S V and S T at right angles to each 
other. Make S V equal to the width of the bracket 
measured horizontally across the face, and upon S T, 
measuring from S, set off the lengths of the several 
dotted lines in the side view, as shown by the small 
figures. From each of points in S T draw lines to V. 
Then these lines will be the real distances between 
points of corresponding number on the lower side of 
the bracket and points of the next higher number upon 



the upper side. The figures in S T correspond with 
the figures upon the lower side of the bracket, the 
point V representing, in the case of each line, the 
next higher number ; thus 2 V is the distance from 2 
to 3' across the face of the bracket, 3 V the distance 
3 4', 4 V the distance 4 5', etc. The dotted lines in 
the side view representing the distances 1 2 and 7 8 
cannot, of course, be shown in that view, because 
they lie in surfaces which ajnpear in profile ; but since 
these surfaces are parallel with the plane of the back 
of the bracket these distances for use in the pattern 
may be taken directly from the front view, as shown 
by the dotted lines 1 2' and 7 8' in that view. 

To lay out the pattern of the face piece first draw 
any line, as C 3 A 3 or 1 1' of the pattern, equal in length 
to 1 1' of the front view. From C 3 of the pattern as 
center, with a radius equal to 1 2' of the front view, de- 
scribe a small arc, which intersect with another arc 
drawn from A 3 as center, with a radius equal to 1' 2' of 
the front view, thus establishing the point 2' of the 
upper side of the pattern of the face. From 2' of the 
pattern as center, with a radius equal to 2 2' of the 
front view, strike a small arc, which intersect with 
another arc struck from 1 of the pattern as center, with 
a radius equal to 1 2 of the front view, thus establish- 
ing the position of the point 2 in the lower side of the 
pattern. From 2 of the pattern as center, with a 
radius equal to 2 V of the diagram of triangles, strike 
a small arc, which intersect with another arc struck 
from 2' of the pattern of center, with a radius equal to 
2' 3' of the side view, thus establishing the point 3' of 
the pattern. From 3' of the pattern as center, with a 
radius equal to 3 3' of the front elevation, strike a 
small arc, which intersect with another arc struck from 
2 of the pattern as center, with a radius equal to 2 3 of 
the side view. So continue, using the distances across 
the face indicated by the dotted lines as found in the 
diagram of triangles in connection with the spaces in 
the profile A 2 B 2 of the side view to form the upper 
side A 3 B 3 of the pattern, and the distances across the 
face as measured upon the solid lines of the front view 
in connection with the spaces upon the profile (7 D 2 to 
form the lower side C 3 D 3 of the pattern, until the 
points 13 and 13' are reached. 

As the lines C A and D B of the front of the 
bracket must be cut to fit the curves of the moldings 
above and below, against which the bracket fits, the 
corresponding lines of the pattern can be drawn with 
radii respectively equal to K E and K F, as shown by 
the curved lines C 3 A s and D 3 B 3 of the pattern. 



Pattern Problems. 

PROBLEM 193. 

Pattern for a Transition Piece to Join Two Round Pipes of Unequal Diameter at an Angle. 



359 



Iu Fig. 626, D C K L shows a portion of the 
larger pipe, of which M P N is the section ; HGBA 
a portion of the smaller pipe, of which E J F I is the 
section ; and ABCD the elevation of the transition 
piece necessary to form a connection between the two 
pipes at the angle HAL. The drawing also shows 
that the ends of the two pipes to be joined are square, 




Fig. 626. — Elevation of a Transition Piece Joining Two Round 
Pipes of Unequal Diameter at an Angle. 

or cut off at right angles, so that the lower base of 
A B C D is a perfect circle whose diameter is D C (or 
M N) and the upper base is a circle whose diameter is 
A B (or E F), and also that the side AD is vertical. 
In the choice of a method of dividing the surface of 
the piece ABCD into triangles, either the elevation 
or the plan can be made use of for that purpose, ac- 
cording to convenience. In the demonstration here 
given the elevation has been used by way of variety, 



all as shown in Fig. 627. Proceed then to divide the 
plan of the upper base into any convenient number of 
equal spaces, as shown, and drop a line from each point 
at right angles to A B, cutting A B, and numbering 
each point to correspond with the number upon the 
plan. In like manner divide the plan of the lower 
base into the same number of equal spaces, and erect 
a perpendicular line from each, cutting the line D G, 
and numbering the points of intersection in the same 
order, or to correspond with the points in the upper 
base, all as shown. Connect the points in D C with 
points of similar number in A B by solid lines, also 




Fig. 627. — Triangulation of Transition Piece Shown in Fig. 626. 

connect points in D C with points of the next higher 
number in A B by dotted lines, which will result in a ■ 
triangulation suitable for the purpose. 

The next step will be to construct sections through 
the piece upon all of the lines upon the elevation (both 
solid and dotted), which operations are shown in Figs. 
628 and 629, and which may be done in the following 
manner : Upon any horizontal line, as T S of Fig. 628, 
erect a perpendicular, as T U. Upon T S set off from 
T the several distances of the points in the lower base 
from the center line I 7 of the plan, as measured upon 
the vertical lines (Fig. 627), all as indicated by the 
small figures. Upon T U set off the lengths of the 



360 



TIi e New Metal Worker Pattern Book. 



solid lines of the elevation, numbering each point thus 
obtained to correspond with its line in the elevation. 
From each of the points upon T U draw horizontal 




W 



\ \J_ 

-\ M— 

\ ,\| 

\3 \j 



x * K\ 






/ v 



\ w 



3 4 



6 



Fig. 628.— Diagram of Sections Fig. 629.— Diagram of Sections 
Taken on Solid Lines of Fig. Taken on Dotted Lines of Fig. 
627. 6 ~?- 

lines to the right, making each in length equal to the 
distance of points of corresponding number in the plan 
of the upper base from the center line 1 7, as measured 
upon the lines at right angles to line A B, thus ob- 
taining the points 2', 3', 4', etc. Now connect these 
points with points of corresponding number in the base 
line T S by means of solid lines, as shown. 

In constructing the sections upon the dotted lines 
of the elevation, shown in Fig. 629, the same course 
is to be pursued as that employed in Fig. 628. The 
base line W V is a duplicate of T S. Upon the per- 
pendicular line erected at W set off the lengths of the 




Fig. 631.— Perspective View of Model. 

several dotted lines of the elevation, numbering each 
point thus obtained to correspond with the number at 
the top of its line in the elevation. From each point 
draw a horizontal line to the right as before, which 
make equal in length to the similar lines in Fig. 628, 
numbering each point as shown by the small figures 



2', 3', 4', etc. Now connect each of these points with 
the point of next lower number in the base line V W 
by a dotted line. 

Having obtained all the necessary measurements, 
the pattern for one-half the envelope of A B C D may 
be developed in the following manner : Draw any line, 
as A D in Fig. 630, which make equal in length to A 
D of the elevation. With A as a center, and a radius 
equal to 1 2 of the plan of the upper base, Fig. 627, 
strike a small arc, which intersect with another struck 
from D as a center, with a radius equal to the dotted 
line 1 2' of the diagram, Fig. 629, thus establishing 
the position of the point 2 in the upper line of the pat- 
tern. From D as a center, with a radius equal to 1 2 
of the plan of the lower base, Fig. 627, strike a small 
arc, which intersect with another struck from point 2, 
just obtained, with a radius equal to the solid line 2 2' 




3 4 5 

Fig. 630.— Pattern of Transition Piece Shown in Fig. 627. 

of the diagram, Fig. 628, thus fixing the position of 
point 2 in the lower line of the pattern. So continue, 
using the lengths of the dotted lines in the diagram, 
Fig. 629, in connection with the lengths of the spaces 
in the plan of the upper base, to develop upper line of 
the pattern, and the lengths of the solid lines in the 
diagram, Fig. 628, in connection with the lengths of 
the spaces in the plan of the lower base, to develop the 
lower line of the pattern, using each combination alter- 
nately until the pattern is complete. As each new 
point of the pattern is determined it should be num- 
bered, and the solid or dotted line used in obtaining 
the same may be drawn across the pattern if desired, 
merely as a means of noting progress, but these lines are 
not necessary, as each point is simply used as a center 
from which to find the next point beyond. 

Sometimes, in order to more thoroughly under- 
stand the method employed in such an operation as the 
foregoing, it is desirable to construct a small model, 



Pattern Problems. 



361 



which can be made from cardboard or thin metal, the 
details of which are clearly shown in Fig. 631. The 
pieces forming the upper and lower bases of it should 
be duplicates of the half plans of the upper and lower 
bases shown in Fig. 627, having the lines there shown 
drawn upon them, and the piece forming the back is a 
duplicate of the plane figure ABCD. These three 
parts may be cut in one piece, after which a right angle 
bend on the lines A B and C D will bring the two bases 
into correct relative position. Five quadrilateral figures 
corresponding to those shown in Fig. 628 may now be 



cut and fastened in position, according to their num- 
bers, between the two bases of the model. Threads 
or wires can be so placed as to correspond in position 
with the dotted lines shown in the elevation, Fig. 027, 
to complete the model. The model is only useful be- 
fore the pattern is developed to assist in showing the 
shapes and order or rotation of the various triangles ; 
and one constructed to the dimensions of any problem 
which may occur to the student at the outset of his 
study of triangulation will serve to assist his imagina- 
tion in all subsequent operations. 



PROBLEM 194. 

The Pattern for a Flaring; Collar the Top and Bottom of which are Round and Placed Obliquely to 

Each Other. 



In Fig. 632 EFGH shows the side elevation of 
a flaring collar, the profile of the small end or top 
being shown at ABCD, and that of the bottom at 
K L M JNT of the plan. The- conditions embodied in 
this problem are in no respect different from those of 
the problem immediately preceding. A slight diffei- 
ence in detail consists in the fact that in the former 
case the short side was at right angles to the larger 
end, while in the present case it is at right angles to 
the smaller end, but the pattern may be obtained by 
exactly the same method as that employed in the 
previous problem. However, as the elevation was 
there made use of to determine the triangulation, the 
plan will here be used upon which to determine the 
position of the triangles of which the pattern will sub- 
sequently be constructed. 

Divide A B C of the profile into any convenient 
number of parts, and, with the T-square at right angles 
to E F of the elevation, carry lines from the points in 
ABC, cutting E F, as shown. Extend the base line 
G H to the left indefinitely, and through the center of 
plan of base draw M, parallel with H G of elevation. 
Drop lines from the points in E F, extending them 
vertically through M With the dividers take the 
distance across the profile A B C D on each of the sev- 
eral lines drawn through it, and set the same distance 
off on corresponding lines drawn through M. That 
is, taking A C as the base of measurement in the one 
case, and M in the other, set off on the latter, on 
each side, the same length as the several lines measure 
on each side of A C. Through the points thus ob- 
tained trace a line, as shown by P Q E, thus obtain- 




ing. 632. — Plan and Elevation of Flaring Collar. 



362 



Hie New Metal Worker Pattern Book. 



ing the shape of the upper outline as it would appear 
in the plan. 

As both halves of the plan when divided by the 
line M are exactly alike, it will only be necessary 
to use one-half in obtaining the pattern ; therefore 
divide K N M of plan into the same number of parts 
as was the half of profile, in the present instance six, 
as shown by the small figures. Number the points in 
K N M to correspond with the points in R Q, and 
connect corresponding points by solid lines, as shown 
by 1 1', 2 2', 3 3', etc. Also connect the points in 
ORQ with those of the next higher number in K N 
M, as shown by the dotted lines 1 2'. 2 3', 3 4', etc. 
The solid and dotted lines thus drawn across the plan 
will represent the bases of a number of right angled 
triangles whose altitudes are equal to the vertical lines 
between E F and J Y of the elevation, and whose hy- 



V W X and on W X set off the lengths of dotted lines 
in plan, and from W on W V the lengths of lines in 
E F Y J of elevation, excepting the line E J, or No. 7, 
which is not used. Connect the points in W V of 
diagram with those of the next higher number in W X, 
as shown by the dotted lines in diagram and by simi- 
lar lines in the plan. The resulting hypothenuses will 
give the correct distances from points in top of article 
to points of next higher number in the plan of 
base. 




2 



*v 



A \ \\ 
\ \ \ \ 



\ 



\ 



T 

Fig. 63 
Based 
Plan i 



7 6 5 4 3 2 i U 

3. — Diagram of Triangles 
upon the Solid Lines of the 
n Fig. 632. 



W 



\ 

\\ 
\ \ 
A^ 



\ \ 



\ 






-, 



6 

63A.- 



J± 



zX 



Fig. 634. — Diagram of Triangles 
Based upon the Dotted Lines of the 
Plan ia Fig. 632. 



pothenuses, when obtained, will give the real distances 
across the sides of the finished article in the direction 
indicated by the lines across the plan. 

To construct these triangles proceed as follows : 
Draw any right angle, as STU in Fig. 633. On T 
U, measuring from T, set off the lengths of solid lines 
in plan, making T U of diagram equal to QM of plan, 
T 2 of diagram equal to 2 2' of plan, T 3 of diagram 
equal to 3 3' of plan, etc. From T on T S set off the 
length of lines in E F Y J of elevation, making T S 
of diagram equal to F Y of elevation, T 2 of diagram 
equal to a 2 of elevation, T 3 of diagram equal to b 3 
of elevation, etc. Connect points in T S with those 
of similar number in T U, as shown by the solid lines. 
The hypothenuses of the triangles thus obtained will 
give the distance from points in plan of base to points 
of similar number of top as if measured on the fin- 
ished article. The diagram of triangles in Fig. 634 is 
constructed in a similar manner. Draw the right angle 




i m 



Fig. 63S. — Pattern of Flaring Collar. 

Having now obtained all the necessary measure- 
ments, the pattern may be developed as follows : Draw 
any line, as q m in Fig. 635, in length equal to S U of 
Fig. 633, or F G of elevation. With m of pattern as 
center, and M 2' of plan as radius, describe a small 
arc (2"), which intersect with one struck from point q 
of pattern as center, and V X of diagram of triangles 
in Fig. 634 as radius, thus establishing the point 2" of 
pattern. With point 2" of pattern as center, and 2 2 
of Fig. 633 as radius, describe a small arc (2), which 



Pattern Problems. 



363 



intersect with another struck from point q of pattern 
as center, and C 2 of profile as radius, thus establish- 
ing the point 2 of pattern. With 2 of pattern as cen- 
ter, and 2 3 of Fig. 634 as radius, describe a small arc 
(3"), which intersect with another struck from 2" of pat- 
tern as center, and 2' 3' of plan as radius, thus estab- 
lishing point 3" of pattern. With 3" of pattern as 
center, and 3 3 of Fig. 633 as radius, describe a 
small arc (3), which intersect with one struck from 
point 2 of pattern as center, and the distance 2 3 of 
profile as radius, thus establishing point 3 of the pat- 



tern. Proceed in this manner, using the hypothenuses 
of the triangles in Figs. 633 and 634 for the distances 
across the pattern ; the distances between the points 
in the plan of base for the stretchout of the bottom of 
pattern; and the distances between the points in the 
profile of top for the stretchout of the top of pattern. 
Lines drawn through the points of intersection, as 
shown by q o and m k, will, with q m and ok, constitute 
the pattern for half the article. The other half of the 
pattern q o' k' m can be obtained by any convenient 
means of duplication. 



PROBLEM 195. 

The Pattern for a Flaring: Flange, Round at the Bottom, the Top to Fit a Round Pipe Passing - through 

an Inclined Roof. 



In Fig. 636, let K L represent the juitch of the 
roof, ABCD the elevation of the flaring flange, A J 
D the half plan of the base, and BBC the half plan 
of round top through which the pipe passes. 




Fig. 636. — Elevation of a Flaring Flange to Fit Against an 
Inclined Roof. 

It will be seen by comparison that this problem 
embodies exactly the same principles as do the two 
immediately preceding, with the slight difference in 
detail that its short side is not at right angles to either 
upper or lower base. Also, in this case the bottom of 
the article appears inclined instead of the top. It will 



be seen at a glance that if the shape be considered as 
anything else than a flange against an inclined roof the 
drawing might be so turned upon the paper as to bring 
the line K L into a horizontal position, when it would 
present the same conditions as those of Problems 193 
and 194 with the slight difference in detail above al- 
luded to. 

The method of triangulation employed in this case 
is exactly the same as in the problem immediately pre- 







jftDG 






v\ 



\ \V\ 



w \ \ \\ 
\ \ \ \ 

\ \ \ 



6 5 



3 2 



Fig. 637.— Diagram of Triangles Fig. 688.— Diagram of Triangles 
Based upon the Solid Lines of Based upon the Dotted Lines of 
the Plan in Fig. 636. the Plan in Fig. 636. 



ceding, and the operation is so clearly indicated by the 
lines and figures upon the four drawings here given as 
scarcely to need explanation, if the previous problem 
has been read. The plans of both top and bottom are 
divided into the same number of equal parts, and a 
view of the top as it would appear when viewed at 
right angles to the base line K L, and as shown by F 
Gr H, is projected into the plan of base, as indicated 
by the lines drawn from B C at right angles to A D. 



364 



The New Metal Worker Pattern Book. 



Points of like number in the two curves FHG and A 
J D are joined by solid lines, and the four-sided figures 
thus obtained are redivided diagonally by dotted lines. 
These solid and dotted lines become the bases of the 
several right angled triangles shown in Figs. 637 and 
638, whose altitudes are equal to the bights given be- 
tween the lines B C and F Gr, and whose hypothenuses 
give correct distances across the pattern between points 
indicated by their numbers. The pattern is developed 
in the usual manner by assuming any straight line, as 
C D in Fig. 439, equal to C D of Fig. 636, as one end 
of the pattern, and then adding one triangle after an- 
other in their numerical order; using the stretchout of 
B E C, Fig. 636, to form the upper line of the pattern, 
the stretchout of A J L to form the lower side of the 
nattern and the various dotted and solid hypothenuses 




Fig. 639. — Pattern of Flaring Flange Shown in Fig. 636. 

in Figs. 637 and 63S alternately to measure the dis- 
tances across the pattern. 



PROBLEM 196. 



Pattern for an Irregular Flaring Article, Elliptical at the Base and Round at the Top, the Top and 

Bottom not Being Parallel. 



The conditions given in this problem are essen- 
tially the same as those of Problem 193, but the fol- 
lowing solution differs from that of the former problem 
in the method of finding the distances from points 
assumed in the base to those of the top, and is intro- 
duced as showing varieties of method : In Fig. 640, 
C G H D represents the side view of the article, of 
which G K H is a half profile of the top and C F D a 
half profile of the base. For convenience in ob- 
taining the pattern the half profiles are so drawn that 
their center lines coincide with the upper and lower 
lines of the elevation. 

Divide both of the half profiles into the same 
number of equal parts — in the present instance eight. 
From the points obtained in the half profiles drop per- 
pendiculars cutting G H and C D. Connect the points 
secured in Gr H with those in C D, as a n, b m, etc. 
Also connect the points in Gr H witli those in C D, as 
indicated by the dotted lines 1 n, a m, etc. Prefer- 
ence to Problem 193 will show that in order to obtain 
the correct lengths represented by the several solid 
and dotted lines drawn across the elevation complete 
sections upon those lines were constructed, as shown 
in Figs. 628 and 629. In the present case these dis- 
tances will be derived from a series of triangles whose 
bases are the differences between the lengths of the 
lines drawn across the half profile of the top and 
those of the bottom. 



Therefore, to obtain the triangles giving the true dis- 
tances represented by the solid lines proceed as follows : 




Fig. 640. — Elevation and Profiles of an Irregular r ,uring Article, 
Showing Method of Triangulation. 

First set off from C D upon each line in the base CFD 
the length of the corresponding line in the top ; thus 



Pattern Problems. 



365 



make n o equal to a 2, mp equal to b 3, I q equal to c 4, 
etc. For the bases of the triangles represented by the 
dotted lines set off from C D the length of correspond- 




»■'«' p' 0' R »' f' 
FKg. 641. — Diagram of Triangles 
Representing Solid Lines of the 
Elevation. 



w 




1 ' 


/, 


a" 


// 




I 


b 


II 




ll 




II 


v c" 


II 


\ 

\ 
\ 


11 


II •• 
in d 
11 


Hi 


\ \ 
\ \ 


,','1 f" 


'l N \ 


1 «" 


J \ \ 


/// / 
in / 


IV \ \ 


/// / 


/// ' 


■ // 1 L 


IV \ .. \ .. 



3"a"X"Z" V 6 '7" 5 " 4 " 
Fig. 64S. — Diagram of Triangles 
Representing Dotted Lines of 
the Elevation. 



ing lines in G K H, as shown by the small figures in 
COD. Thus make m 2' equal to a 2, I 3' equal to 
b 3, h i' equal to c 4, etc. The triangles represented 
by solid lines in the elevation, and shown in Fig. 6-41, 
are obtained as follows : Draw the line P Q, and from 



from C to are set off to the left of R, and the lengths 
from O to P on R Q. Thus make R a' of diagram 
equal to n a of Fig. 640, R 0' of diagram equal to o 
17 of half profile of base, and connect a! a'. Make R 
b' of diagram equal to m b of Fig. 640, R p' of diagram 
equal to p 16 of the base, and connect p' 6', etc. 

The triangles represented by clotted lines in C G 
H P are obtained in a similar manner. Praw the line 
T U in Fig. 642 and erect the perpendicular V W. 
From V, on V W, set off the lengths of dotted lines 
in C Gr H P of the elevation. Thus make VI" equal 
to n 1 of Fig. 640, Y a" equal to m a, V b" equal to 

1 b, etc. Upon V T or V U set off the lengths of the 
lines in C P F of Fig. 640, as indicated by the 
small figures. Thus make Y X" equal to n 17 of the 
base in Fig. 640, and draw X" 1". Make V 2" equal 
to 2' 16 of the base, and draw 2" a", etc. 

To obtain the pattern first draw the line C G of 
Fig. 643, in length equal to C G of Fig. 640. From 
C, with radius equal to C 17 of the half profile of 
base, strike a small arc (0), which intersect with an- 
other arc struck from G as center, and 1" X'' of Fig. 
642 as radius, thus establishing point in the curve 
of the pattern. From G of pattern as center, and G 

2 of the half profile of top as radius, strike a small 
arc, which intersect with another arc struck from 




Fig. 643. — Pattern of Shape Shown in Fig. 640. 



a convenient point erect the perpendicular R S. From 
R on R S set off the lengths of solid lines in C G H P, 
and from R on P Q set off the lengths of correspond- 
ing lines in C O P F, as indicated by the small letters 
in C P. To avoid a confusion of lines the lengths 



o of pattern as center, and a' 0' of Fig. 641 as 
radius, thus establishing point a in the upper curve of 
the pattern. From of pattern as center, and 17 16 
in F as radius, strike a small arc (p), which inter- 
sect with another arc struck from a of pattern as cen- 



366 



The New Metal Worker Pattern Book. 



ter, and a" 2" of Fig. 642 as radius, thus locating 
point p of the pattern. From a of the pattern as cen- 
ter, and 2 3 in G K as radius, strike a small arc, 
which intersect with another arc struck from p of pat- 
tern as center, and p V of Fig. 641 as radius, thus 
locating point b of pattern. Proceed in this manner, 
using in the order named the spaces in the lower pro- 
file, the hypothenuses of triangles in Fig. 642, the 



spaces in the upper profile, and the hypothenuses of 
triangles in Fig. 641. The points thus obtained, as 
indicated by the letters in Fig. 643, are the points 
through which the pattern lines are to be traced. Then 
C G H D is the required pattern for one-half the 
article. The other half of pattern, as shown by C G H' 
D', can be obtained by any convenient method of du- 
plication. 



PROBLEM 197. 

The Patterns for a Bathtub. 



In Fig. 644, let A B C D be the elevation and E 
F G H half the plan of a bathtub. An inspection of 
the drawing shows that neither the segments forming 



irregular in character, and the only available method 
by which the various dimensions and curves constitut- 
ing the patterns of the same can be ascertained is by 










70/11 p 

532 
TRIANGLES IN 
HEAD PIECE 



— ^^4^ 



7ig. 644. — Plan and Elevation of Bathtub, with Diagrams of Triangles, Section of Top and Pattern of Side Piece. 



the . head nor those of the foot of the tub are concen- 
tric, and that their upper and lower bases are not par- 
allel. Therefore the figures which they constitute are 



dividing their surfaces into small triangles, which can 
most easily be accomplished in the following manner: 
Divide each of the curves J I and G H, forming the 



Pattern Problems. 



iHil 



plan of the head piece, into the same number of equal 
parts, numbering each the same, as shown, and con- 
nect points of similar number by solid lines. Also 
connect each point in J I, the line of the bottom, with 
the point of next higher number in G H, the line of the 
top, by a dotted line, all as shown. The curves E F 
and L K, forming the plan of the foot piece, are also to 
be divided into spaces and the points connected by 
solid and dotted lines in the same manner as those of 
the head. 

The solid and dotted lines thus drawn between the 
points in the two curves of the plan will form the 
bases of a series of right-angled triangles, whose hypoth- 
enuses (after the altitudes are obtained) will give the 
real distance between the points whose number they 
bear upon the finished article. As, owing to the slant 
of the top line A B of the elevation, the triangles will 





Fig. 645.— Half Pattern of 
Head Piece. 



Fig. C4C.— Half Pattern of 
Foot Piece. 



be of differing bights, the simplest way of constructing 
them will be as follows : Upon D C of the elevation 
extended, as a base, erect a perpendicular line, M N 
From N on the base line set off the various lengths of 
the solid lines in the plan of the head piece, as shown 
toward Q. From each of the points in the curve G H 
erect perpendicular lines, cutting A B of the eleva- 
tion ; and from these points of intersection carry lines 
horizontally to the right, cutting the line M N, num- 
bering each point to correspond with the points in G H, 
all as shown. Lines connecting j)oints of similar num- 
ber at M and Q will be the hypothenuses required, or 
the real distances between points of similar number in 
the top and bottom of the finished article. In a sim- 
ilar manner erect another perpendicular, P, and set 
off from P on the base line the lengths of the several 
dotted lines in the plan of the head piece, as shown 
toward R. The bights of the points in the curve of 
the top can be determined upon the line P by con- 



tinuing the lines drawn from A B toward M till they 
intersect P. Each point in the base P E is now to 
be connected with the point of the next higher number 
in P by a dotted line. The various hypothenuses 
drawn between and R will then be the correct dis- 
tances between the points connected by the dotted lines 
of the plan. (The numbers in P B, correspond with 
those in I J of the plan and not with H Gr). The dia- 
grams from which the dimensions for the foot piece are 
obtained are shown at S T U and V W X at the left 
of the elevation and are obtained in a manner exactly 
similar to those just obtained for the head piece, 
all of which is clearly shown by the lines of the 
drawing. 

From an inspection of the drawing it will be seen 
that the line E F G H does not represent the true 
lengths or measurements taken on the top line of the 
tub, because A B, not being horizontal, is longer than 
the line E H, its equivalent in the plan, and therefore 
a true section on the line A B must be obtained, as 
shown above the elevation. This may be accomplished 
in the following manner : At any convenient distance 
above A B draw E 1 H 1 parallel to A B, and from all 
of the points previously obtained on A B carry lines 
at right angles to A B, cutting E 1 H 1 , and extend them 
beyond indefinitely, numbering each line to correspond 
with the point in E F Gr H from which it is derived. 
On the line 2 set off from E 1 H' a distance equal to 
the distance of point 2 of the plan from the line E H 
as measured on the vertical line; on line 3 set off as 
before a distance equal to the distance of point 3 of 
the plan from E H, and so continue until the distances 
from E H of all the points in E F G H have been 
transferred in like manner to the new section. Then 
a line traced through these points, as shown by E 1 F' 
G 1 H', will be a section or plan on the line A B, from 
which measurements can be taken in developing the 
upper edge of the pattern. 

Having obtained, by means of the various dia- 
grams constructed in connection with the elevation, the 
correct distances between all the points originally as- 
sumed in the plan, the pattern may now be developed 
by simply rejDroducing all of these distances or meas- 
urements in the order in which they occur upon the 
plan. For the pattern of the head piece assume any 
line, as I H of Fig. 645, which make equal in length to 
C B of the elevation, or what is the same thing, equal 
to 1 1 of the diagram MNQ. From H as a center, 
with a radius equal to 1 2 of the section of top, strike 
a small arc, which intersect with another arc struek 



368 



Tlie New Metal Worker Pattern Book. 



from I as a center, with a radius equal to 1 2 of the 
diagram of dotted lines OPE, thus establishing the 
point 2' of the pattern. From 2' as center, with a 
radius equal to 2 2 of the diagram of solid lines M N Q, 
strike a small arc, which intersect with another struck 
from point I as a center, with a radius equal to 1 2 of 
the line 1 -1 of the plan, thus establishing the poinl 2 
of the pattern. Continue this operation, using in nu- 
merical order the distances taken from the top section, 
in connection with the distances obtained from the 
diagram of dotted lines P E, to form the top line of 
the pattern, and the distances taken from the diagram 
of solid lines M N 0, in connection with the distances 
measured upon the bottom line I J of the plan, to form 
the bottom line of the pattern, all as indicated by the 
solid and dotted lines drawn across a portion of the 
pattern. Then IHGJ will be one-half the pattern of 
the head piece. The pattern for the foot piece is de- 
veloped in exactly the same manner by making E L 
equal to A D of the elevation, and using the diagram 
of dotted lines V W X to measure upon the pattern 
the distances indicated by the dotted lines upon the 
plan, and the diagram of solid lines ST U to measure 
upon the pattern the distances indicated by the solid 



lines across the plan, the distances forming the top 
line of the pattern being taken from E 1 F' of the top 
section while the distances forming the bottom line of 
the pattern are taken from the line L K of the plan. 

The pattern for the flat portion of the side F G J K 
can be obtained as follows : Parallel to K J of the 
plan draw any line, as K 1 J 1 . At right angles to K J 
of the plan project lines from points K, J, F and G, 
cutting K 1 J 1 , as shown, establishing the points K 1 and 
J 1 , and continuing the lines from points F and G in- 
definitely. From K 1 of the pattern as a center, with a 
radius equal to 8 8 of the diagram S T U, or of Fig. 
64-1, strike a small arc, cutting the line projected from 
point F of the plan, as shown at F 2 of the pattern. 
From J 1 of the pattern as a center, with a radius equal 
to 7 7 of the diagram M N Q, or of Fig. 644, strike a 
small arc, cutting the line projected from the point G 
of the plan, as shown at G 2 . Draw the lines K 1 F 2 , 
F 2 G 2 and G 2 J 1 ; then K' F 2 G 2 J 1 will be the pattern of 
the flat portion of the side. 

The patterns of the several parts can be joined 
together, by any convenient method of duplication, in 
such a manner as to produce as much of the entire 
pattern in one piece as it is desired. 



PROBLEM 198. 

The Pattern for a Flaring Flange to Fit a Round Pipe Passing through an Inclined Roof; the Flange 
to Have an Equal Projection from the Pipe on All Sides. 



In Fig. 647, let a b c d be the elevation of the 
pipe, E E' its plan, A B C D the elevation of the 
flange and C D the angle or pitch of the roof. Since 
the projection of the base of the flange is required to 
be equal on all sides, as shown by C 1 and D 1, the 
flange will appear in the plan as a perfect circle, F F'. 
To avoid confusion of lines another elevation of the 
flange G H K J is shown in Fig. 648, below which is 
drawn a half plan of its base, MBN, and above which 
is a half plan of its top, G L H, all of which will be 
made use of in dividing the surface of the flange into 
measurable triangles for the purpose of developing a 
correct pattern of the same. 

Divide the semicircle G L H into any convenient 
number of equal parts— in the present instance 12 — 
and from the points thus obtained drop perpendicular 



lines to G H. To obtain the shape of section on roof 
line J K divide the half plan of base MBN into the 
same number of equal parts as was G L H, and from 
the points thus obtained carry lines at right angles to 
M N, cutting J K. From the points in J K draw 
lines at right angles to it, as shown by a 1, b 2, c 3, 
etc. On these lines, measuring from J K, set off the 
length of corresponding lines in M N B, thus making 
lines a 1, b 2, c 3, etc., in JKC equal to lines al, b2, 
c 3, etc., in M N B. A line traced through these 
points, as shown by J C K, will give the shape of sec- 
tion on roof and furnish the stretchout of base for 
obtaining the pattern. 

In Fig. 649 is drawn a duplicate of the plan in 
Fig. 647, the spaces in its outer line D P being 
exact duplicates of the spaces in M B N of Fig. 648, 



Pattern Problems. 



369 



and the spaces in its inner line 0' D' P' being dupli- 
cates of those in G L H, all as shown by the small 
figures. Draw solid lines connecting similar points, 
as 1' 1, 2' 2, 3' 3, etc. In like manner connect the 




PLAN 
Fig. 647. — Plan and Elevation of Pipe and Flange. 



points in 0' D' P' with those of the next higher num- 
ber in ODP, as with 1', 1 with 2', 2 with 3', etc., 
with dotted lines. These solid and dotted lines will 



then form the bases of a series of right angled trian- 
gles, whose altitudes can be derived from the eleva- 
tion, and whose hypothenuses, when obtained, will be 
the correct measurements across the pattern between 
points of numbers corresponding with the lines across 
the plan. 

To construct the diagrams of triangles represented 
by solid and dotted lines in plan, extend G H of Fig. 
648 indefinitely, as shown by H W. From the points 
in J K carry lines to the right indefinitely, parallel 
with G W, as shown by the lines between G W and 
K Y. At any convenient place, as R, and at right 
angles to GW, erect the line R S, cutting the base line 
K Y. From R set off the distance R T, equal to the 
length of any of the solid lines in plan, Fig. 649, as 
P' P, which is the horizontal distance between the 
pipe and lower edge of the flange. Draw T U parallel 
with R S, and also draw lines from the points in R S 
to T. For convenience the points in R S can be num- 
bered to correspond with the points in J C K. Then 
the triangle TUS will correspond to a section through 




Wig. 649. — Plan of Flange, Showing Triangxdation. 

the article on the line P' P in plan, the hypothenuse 
S T representing the distance between the pipe and 
lower edge of the flange. The diagram of triangles in 

V W Y X is constructed in a similar manner; draw 
W Y at right angles to G W, and set off the space W 

V equal to the length of one of the dotted lines in 
plan, Fig. 649, as 1', and draw lines from the points 
in W Y to V. 

In developing the pattern the stretchout of top of 



370 



T)ie New Metal Worker Pattern Book. 



can be obtained from 
stretchout of lower edge 



flange where it joins the pipe 
the semicircle G L H. The 

of flange where it joins the roof can be obtained from 
the section on the roof line J C K. The distance be- 
tween points in D P and 0' D' P' of plan, Fig. 649, 



length to T S of first diagram of triangles. With the 
dividers set to the distance K 1 in K C J of section 
strike a small arc (1') from the point 0' of pattern. 
With the dividers set to the distance Y 1 of second dia- 
gram of triangles strike a small arc from the point of 




SECOND 
DIAGRAM 



Fig. 6 J/S.— Elevation of Flange, with Plan, Sections and Diagrams of Triangles. 



as indicated by the solid lines, is given in the diagram 
of triangles T E S. The distance between points 
as indicated by dotted lines in plan is given in the 
diagram V W Y. For the pattern then proceed as 
follows : Draw any line, as H' K', Fig. 650, equal in 



pattern as center, cutting the first arc at 1' of pattern. 
From point 1' of pattern as center, and T 1 of first 
diagram of triangles as radius, describe a small arc (1), 
which intersect with one struck from of pattern as 
center, and 1 in H L G as radius. Thus the points 



Pattern Problems. 



371 



0' and 1 1' of pattern are established. Proceed in 
this manner, using in the order described the stretch- 
out obtained from the elliptical section K C J, the hy- 
pothenuse of triangle in second diagram corresponding 
to the dotted line in plan, the stretchout from the sec- 
tion GLH, and the hypothenuse of triangle in the 



other half of the pattern can be obtained by any means 
of duplication most convenient. 

Should it be required to construct such a flange 
to fit over the ridge of a roof, it is clear that that part 
of the flange shown in the plan, Fig. 649, by OO'D'D 
would be a duplicate of the part shown by D' P' P D, 



•' 12 




J '2' 



Fig. 650.— Half Pattern of Flange Shown in Fig. 647. 



first diagram corresponding to the solid line drawn 
across the plan, until all the measurements are used. 
Lines traced through the points thus obtained, as shown 
by H' G' and K' J', will be the half pattern. The 



and that, therefore, that portion of the pattern, Fig. 
650, shown by 6 H' K' 6' would be one-quarter of the 
entire pattern, which could be duplicated so as to 
make either one-half or the whole pattern in one piece. 



PROBLEM 199. 



Pattern for the Hood of a Portable Forge. 



In Fig. 651, C A B D represents the front eleva- 
tion of a hood such as is frequently used upon a por- 
table forge, K L M N its plan and EFHJ a its side 
view. The opening A B at the top of the hood is 
round, as shown by L P of the plan, while the base C D 
where it joins the forge is nearly semi-elliptical, as 
shown by K L M of the plan. In the side elevation 
E a shows the amount of flare and projection of the 



front of the hood, while the opening, shown in the front 
by C S D, appears as a simple straight line, a J. With 
these conditions given, the arch of the opening C S D 
of the front elevation can be determined in connection 
with the plan, by projection, as shown by the hori- 
zontal dotted lines, while if the arch of the front eleva- 
tion be assumed arbitrarily then its line (a J) in the side 
view must be obtained by projection, and will be either 



372 



The New Metal Worker Pattern Book. 



straight or curved according to the nature of the curve 
employed in the front elevation. 

Assuming the straight line a J of the side view as 
the true profile of the arch, its curve in either the front 
elevation or the plan must be determined, as a means 




FRJONT ELEVATION 

L 



91011 




SIDE ELEVATION 



Fig. 651. — Front and Side Elevations and Plan of Hood, Showing 
System of Triangulation. 



of obtaining the pattern. As the flaring portion of the 
hood very much resembles a conical frustum having 
an oblique base, probably the simplest method of ar- 
riving at its true shape is to first determine the plan of 
this irregular frustum of which it is a part. Therefore 
produce the oblique line E a of the side elevation until 
it intersects the base line H J extended in the point 
G. Next set off from L on the center line of the plan 
a distance equal to H Gr of the side elevation, thus 
locating the point R. Through R, from a center to be 
determined upon the center line, draw the curve form- 



ing the front of the plaD., with such length of radius as 
will make an easy junction with the curves of the back 
at K and M. It is not necessary that the curve K R 
M should be a perfect circle throughout; it may 
change as it approaches K and M so as So flow 

smoothly into the as- 
sumed curve of the 
back. It is simply 
necessary that no 
angle be produced at 
K and M, as such an 
angle would be con- 
tinued through the 
surface of the hood 
toward the opening 
of the top. 

Divide the circle 
of the top P L into 
any convenient num- 
ber of equal spaces, 
as shown by the small 
figures ; also divide 
the outer curve of the 
plan EML into the same number of spaces. For accu- 
racy and convenience it will be found advisable to make 
the spaces shorter as the curve increases from M toward 
L until the end of the curve is reached at the point 11. 
Connect points of similar number in the two curves by 
solid lines, as shown; also connect points in the plan 
of the top with points of the next higher number in 
the plan of the base by dotted lines. In order to pro- 
duce the curve of the opening correctly in the plan 
and the front elevation it will be necessary first to 
draw upon the side elevation lines corresponding to 
the solid lines just drawn across the plan. To accom- 
plish this place the "|"-square at right angles to L R of 
the plan, and, bringing it successively against the points 
in the plan of the base R M L, drop corresponding 
points on L R, as shown. Transfer the spaces thus 
produced to the base line H Gr of the side elevation, 
numbering each point to correspond with the plan. 
By means of the J-square placed as before, drop points 
from the plan of the top to the center line L P (omitted 
in the drawing to avoid confusion of lines) and transfer 
the same to the line FE of the side elevation, number- 
ing each point as before. Now connect points of cor- 
responding number in the upper and lower lines of the 
side elevation by solid lines, as shown ; then will these 
lines be the elevations of the solid lines drawn across 
the plan. 



Pattern Problems. 



373 



It may be here remarked that, as the pattern will 
be obtained from the plan, a correct front elevation of 
the opening, or arch, is not necessary to the work, but 
if it is desired it can be obtained in the following man- 
ner : Place the T-square parallel to L R of the plan 
and, bringing it against the ]3oints in the plan of the 





*7///'iii 
'// / / //iii 



B Ffr~. 



//// / ' 



//III 



..'_ 



_2 L 



6 9 10 II 



Fig. C5S. — Diagrams of Triangles. 



base between R and M, drop corresponding points on 
the base line C D of the front elevation. Also in the 
same manner drop points from the curve of the top L P 
in plan upon A B of the front elevation, and connect 
points of corresponding number in the two lines by 
solid lines, as shown. From the points of intersection of 
the solid lines in the side elevation with the line a J (the 
profile of the arch), a, b, c, etc. , carry lines horizontally 
across, as shown, intersecting them with lines of cor- 
responding number in the front elevation. A line 
traced through the points of intersection as shown from 
S to D, will be the correct elevation of the opening in 
the front of the hood. 

The correct plan of the opening may be obtained by 
placing the T-square parallel to L E and bringing it 
against the various points of intersection through which 
the curve S D was traced and cutting the solid lines 
of corresponding number in the plan, giving the points 

a, b, c, etc. In case the development of the curve S D 
has been omitted, measure the horizontal distance of 
each of the points a, b, c, etc., in a J of the side eleva- 
tion from the line F H and set off the same on the 
center line of the plan from L toward N". Thus the 
horizontal distance of point a from the line F H is set 
off from L on the center line of the plan, thus locating 
the point N or a, the extreme point of projection of the 
hood. In the same manner the projections of points 

b, c, etc., of the side elevation, or in other words, their 
distances from F H are set off from L of the plan, as 



shown between N and T. Now place the T-square at 
right angles to L R and, bringing it against these points 
last obtained, cut the corresponding solid lines of the 
plan, thus locating the points a, b, c, etc., of the plan, 
as before. A line traced through these points will be 
the correct plan of the curve of the opening. 

Before the pattern can be begun it will be neces- 
sary to first obtain the correct distances represented by 
the solid and dotted lines across the plan. This is ac- 
complished by means of two diagrams of triangles, 
shown in Fig. 652, as follows: Draw the vertical line 
A B, in length corresponding to the hight of the hood, 
as indicated by F H in the side elevation. At right 
angles to A B draw B C, in length corresponding to P 
R or 1 1 of the solid lines of the plan. From B set off 
also the spaces B 2, B 3, B 4, etc., corresponding in 
length to the lines 2 2, 3 3, I 4, etc., of the plan. Con- 
nect the points in B C with the point A by solid lines. 
Then will these lines represent the true distances between 
points 1 and 1, 2 and 2, etc., of the plan. The second 
diagram of triangles is constructed in a similar manner. 
The vertical line E D is drawn, equal to F H of the side 
elevation. E F is set off at right angles to it, in length 
equal to the dotted line 1 2 of the plan. From E are 
set off the distances E 3, E 4, etc., corresponding to 
the lines 2 3, 3 4, etc., of the plan. The points thus 
established in F E are then connected with D by means 
of dotted lines. Then will these lines represent the 




6 --/io^s " m 

Fig. 653.— Pattern of Hood. 



true distances between points 1 and 2, 2 and 3, etc., 
of the plan. 

To develop the pattern, first draw any vertical line, 
as L Z of Fig. 653, representing the center of the 



374 



Tlie New Metal Worker Pattern Book. 



back, which make equal to the hight of the hood F H. 
As the base of the hood is perfectly straight from L to 
the point 11, set off on a horizontal line from the point 
Z, in Fig. 653, a distance equal to L 11 of the plan, 
and draw 11 L of the pattern. With L as center, and 
11 10 of the small circle in plan as radius, describe a 
short arc. Then, from 11 of the base in pattern as 
center, and 11 D of the second diagram of triangles as 
radius, describe a short arc intersecting the one first 
drawn, thus establishing the point 10 of the ripper line 
of the pattern. Then from this point as center, with 
A 10 of the first set of triangles as radius, describe a 
short arc, and from 11 of the base of the triangular 
portion of the pattern, with 11 10 of the outer curve 
of the plan as radius, describe another arc intersecting 
it, thus establishing the point 10 in the lower line of 
the pattern. Proceed in this manner, using alternately 
the spaces in the inner line of the plan, the hypothe- 
nuses of the dotted triangles, the hypothenuses of the 
triangles indicated by solid lines, and the spaces in the 
outer line of the plan, obtaining the several points, as 
shown. Then lines traced through these points will 
be the pattern of the envelope of the shape indicated 
by F E G H of the side elevation, or in other words, 
of the frustum of which the hood forms a part. It now 



remains to cut away such a portion of this pattern as 
rejoresents the part G a J of the side elevation. To ac- 
complish this it is simply necessary to obtain the posi- 
tions of the points a, b, c, etc., of the plan and side 
elevation upon the lines 1 1, 2 2, 3 3, etc., of the pat- 
tern. 

With the blade of the T-square set parallel to the 
base line G H of the side elevation brine it against the 
points of intersection made by the line a J with the 
radial lines, and cut the vertical line F H, as shown by 
the short dashes drawn through it. Transfer the points 
thus obtained in F H to the vertical line A B of the 
first set of triangles. Then with the blade of the T- 
square at right angles to A B, and brought successively 
against the points in it, cut the hypothenuses of the 
several triangles corresponding in number to the lines 
from which the points were derived in the side eleva- 
tion, all as indicated by the letters a, b, c, d, e and/. 
The distances of these points from A may now be 
transferred to lines of corresponding number in the 
pattern, measuring from the upper line, as shown by 
a, b, c, etc. Then a line traced through these points, 
as shown from 1ST to M, will give the shape of the front 
or arch of the hood, and LPS M Z will be the half 
pattern of the hood. 



PROBLEM 200. 



The Patterns for the Hood of an Oil Tank. 



In Fig. 654 are shown the elevations and plan of a 
hood of a style which is usually hinged to the top of 
an oil tank, or can. The plan shows a curve of some- 
thing more than a semicircle, H' G F', while the curve 
F K H of the back view is slightly less than a half 
circle, the problem being to determine the shape of a 
piece of metal to fill the space between the two curves, 
as shown by A B C of the side view. 

Divide one-half of the plan into any number of 
equal parts, as shown by the small figures 1, 2, 3, etc. 
From the points established in the plan carry lines up- 
ward until they cut the base line of the required piece, 
as indicated by the points between A and B. From 
the points thus established carry lines parallel to A C 
until they cut the line representing the back of the 
hood, as shown between C and B, thence carry them 



horizontally until they cut the profile of the back of the 
hood, as shown by the points between K and F. From 
the points in K F drop lines vertically on to the base 
line F E, establishing points in it, as shown. Lay off 
spaces in the line F' E' of the plan corresponding to 
those of F E in the back, and from the points thus 
established draw solid lines to those of corresponding 
numbers laid off in the plan from G to F'. These 
lines represent the bases of a series of right angled 
triangles whose altitudes are shown by the dotted lines 
of the back view, and whose hypothenuses will give 
the correct distances between points of similar number 
in the plan. 

As the altitudes of these triangles are also shown 
in C B of the side elevation, that view is here made 
use of for the purpose of obtaining the required hy- 



Pattern Problems. 



375 



pothenuses. However, since the solid lines drawn 
across the plan are not parallel to Gr E', the distances 
1 B, 2 B, etc., representing them in the base line of the 




Fig. 654. — Elevations and Plan of Hood for an Oil Tank, Shoii'ing 
System of Triangulation. 

side view will not be the correct bases of the triangles, 
therefore set off on A B, measuring each time from B, 
the correct lengths of the several solid lines of the plan, 
as indicated by the points near 1, 2, 3, etc., on the line 




Fig. 655. — Diagram of Triangles Based upon the Dotted 
Lines of the Plan. 

A B, from which points draw lines (shown dotted) to 
points of similar number in B C. Then the dotted 
lines 11, 2 2, etc., of the side view will be the correct 
hypothenuses of the triangles whose bases are indicated 



by the solid lines drawn across the plan. To com- 
plete the measurements necessary for obtaining the 
pattern connect the points in the opposite sides of the 
plan diagonally, as, for example, of 
the front and 1 of the back, and 1 of 
the front with 2 of the back, as shown 
by the dotted lines. These clotted 
lines represent the bases of a second 
set of triangles, to be constructed in 
the same manner as the former set, all 
as shown, Fig. 655. Draw A B andB 
C at right angles to each other and 
upon C B set off the several bights 
shown in G B of Fig. 654. Upon A 
B lay off B 0, corresponding in 
length to 1 in the plan. Make B 1 
diagram equal to 2 1 of the plan, and in the 
make B 2 and B 3 of the diagram equal 
to 3 2 and 4 3 of the plan respectively. From the 
points thus established in the base line of the diagram 



same manner 

Q 




Fig. 656.— Pattern for the Top of Hood. 

draw lines to points of next higher number in the ver- 
tical line. These hypothenuses will then represent 
lengths of lines measured on the face of the hood 
corresponding to the diagonal dotted lines in the plan. 
To develop the pattern, first draw any line, as 
of Fig. 656, equal in length to A C of side, Fig. 
654. From 0, at the right of the pattern, as center, 
with the distance between the points to 1 in the pro- 
file F K of the back as radius, describe a short arc. 



376 



Tlie New Metal Worker Pattern Book. 



Next take in the dividers the distance 1 of Fig. 655, 
and from the opposite end of the center line describe 
a short arc, intersecting the one already drawn at the 
point 1, thus establishing that point. From i as 
center, with dotted line 1 1 of the side view as radius, 
describe another short arc, which in turn intersect by 
an arc struck from of the left hand side of the pat- 
tern with 1 of the plan as radius. This will estab- 
lish the point 1 of the opposite side of the pattern. 
Continue in this way, intersecting the hypothenuse of 
the triangles whose bases are the dotted lines of the 
plan with the measurements taken from the back view, 



and the hypothenuse of the triangles which are shown 
by the solid lines of the j}lan with the measurements 
taken from the circumference of the plan. In this 
manner all points in the profile of the pattern neces- 
sary to its delineation will be established. A free- 
hand line drawn through these points will give one- 
half the required pattern, all as shown in Fig. 656. 
The other half may be obtained by any convenient 
method of duplication. 

The shape of patterns forming the back and the 
vertical sides of the hood are clearly shown in the 
engraving and need no further explanation. 



PROBLEM 201. 

Pattern for an Irregular Flaring Shape Forming: a Transition from a Round Horizontal Base to a 

Round Top Placed Vertically. 



In Fig. 657, let I D E F H represent the front 
elevation of the article, showing the circular opening 
DEFG forming its upper perim- 
eter or profile. The triangle A B 
C shows the shape of the article as 
it appears when viewed from the 
side, below which is drawn the plan, 
showing its circular base, J K L M. 
The line N P shows the plan of the 
opening DEFG, which opening is 
shown in the side view by that por- 
tion of the line A B from A to Q. 
Opposite the front side of the plan 
N P is drawn a duplicate of the pro- 
file D E F G, as shown by E' F' G', 
so placed that its vertical center line 
E' G' shall coincide with the center 
line of the plan, as shown. As the 
article consists of two symmetrical 
halves it will only be necessary to 
develop one-half the complete pat- 
tern. Therefore divide one-half of 
both profiles E F G and E' F' G' into 
the same number of equal parts, 
numberino- each in the same order, 
as shown by the small figures ; also 
divide the plan of the base into the 
same number of equal parts as the 
profile, numbering the points to 
correspond with the same. Drop lines from the points 
on the profile E' F' G' on to the line N P, at right an- 
gles to the same, as shown, and connect these points 



with those of similar number upon the plan of base by 
solid lines, as shown. Also connect points upon the 




i M 



Fig. 657. — Elevations and Plan of an Irregular Shape Forming a Transition from 
a Round Horizontal Base to a Bound Top Placed Vertically. 



base with those of the next higher number upon the 
line N P by dotted lines. 

It will be noticed that the point J of the plan 



Pattern Problems. 



377 



represents at once the point 1 of the base and the 
points 1 and 7 of the profile, shown by B, Q and A 
of the side elevation. The lines drawn across the 
plan represent the horizontal distances between the 




fi 5 4 3 \ -i q 2 3 4 

Fig. 658. — Diagram of Triangles. 

points which they connect and will form the bases of 
a series of right angled triangles, whose altitudes can 
be derived from the elevations, as will be shown, and 
whose hypothenuses, when drawn, will give the true 
distances between points of corresponding number 
across the finished article or its pattern. To obtain 
the altitudes of the triangles carry lines from the 
points in the half profile E F G horizontally 
across, cutting the line A Q, as shown ; then 
the distances of the points in A Q from B 
will constitute the respective altitudes of the 
triangles. Therefore, to construct a diagram 
of all the triangles, draw any horizontal line, 
as D C, Fig. 658, near the center of which 
erect a perpendicular, B A. Upon B A set 
off from B the various distances from B to 



spaces forming the upper or shorter side of pattern can 
be measured from either of the profiles. 

To develop the pattern it is simply necessary to 
construct the various triangles whose dimensions have 
been obtained in the previous operations, 
beo-innina; at either end most convenient 
and using the dimensions in the order in 
which they occur until all have been used 
and the pattern is complete. 

Therefore, upon any straight line, as 
A C of Fig. 659, set off a distance equal 
to A C of Ficr. 657 or the solid line 7 7 
of Fig. 658. From C as a center, with 
a radius equal to 7 6 of the plan, Fig. 657, 
describe a small arc to the left, which 
intersect with another small arc struck 
from A as a center, and with a radius 
equal to the dotted line 7 6 of the diagram 
of triangles, Fig. 658, thus establishing the position 
of the point 6 of the pattern. From A of Fig. 659 as 
center, with a radius equal to 7 6 of the profile, Fig. 
657, describe a small arc, which intersect with another 
struck from point 6 of pattern as center, with a radius 
equal to the solid line 6 6 of Fig. 2, thus locating the 
position of the point 6' of the pattern. 



a r 



points in A Q of Fig. 657. number 



the 



same as shown by the small figures. From 
B set off on B C the lengths of the various 
solid lines drawn across the plan, Fig. 657, 
and connect points in B C with those of 
like number in B A. From B set off 
toward D the lengths of the various dotted 
lines drawn across the plan and connect 
them by dotted lines with points of the next 
higher number in the line B A, all as shown ; 
then these various hyjDothenuses will constitute the 
true distances across the finished article between 
points of corresponding number indicated on the plan 
and elevations. The distances between points in the 
base line forming the larger or outer curve of the pat- 
tern can be measured from the base line in plan, while 




Fig. 659. — Pattern of Shape Shown in Fig. 657. 

So continue to use the spaces of the plan, the 
lengths of the dotted lines of the diagram of triangles, 
the spaces in the profile and the lengths of the solid 
lines of the diagram of triangles in the order named 
until all have been used and the pattern is complete. 
Lines traced through the numbered points obtained, as 



378 



Tlie New Metal Worker Pattern Book. 



shown from C to B and from A to Q, will form the out- 
lines of the pattern for half the article. The other half, 



A Q' B' C, can be obtained by any means of duplica- 
tion most convenient. 



PROBLEM 202. 



Pattern for the Lining of the Head of a Bathtub. 



In Fig. 660 are shown a plan and side and end 
views of the head of a bathtub or the lining of a tub 
the body of which is constructed of wood. The end 
view shows the bottom corners of the tub to be rounded, 
as shown at C G 1 and B 1 F 1 ; the plan shows the head 



any convenient number of equal spaces, as shown by 
the small figures. In like manner divide the quarter 
circle B 1 F 1 of the end view into the same number of 
equal spaces, less one, as also indicated by the small 
figures. From the points thus established in B 1 F 1 carry 




Fig. 660. — Elevations and PUxa of Lining for a Bathtub, Showing 
Ti'iangulation of the Head Piece. 

to be semicircular, while the side view shows that 
the junction between the head and the sides is 
made on the vertical line B 2 W. It will thus be seen 
that the conditions here given are the same as in the 
previous problem — viz. , an irregular flaring piece form- 
ing a transition between two quarter circles (instead of 
complete circles as in the previous problem) lying in 
planes at right angles to each other. 

Divide the quarter circle A F of the top view into 



lines to the horizontal line B F in the top view and 
mark the intersection by small figures, as shown. The 
reason for using one less space in the quarter circle 
B' F' than in the large arc A F is because B' F 1 is not 
the complete profile of the end which is to be connected 
with A F of the top ; the line F 1 E being required to 
complete the same, thus constituting the remaining 
space. Having established these two sets of points in 
the plan, connect those of like numbers, as 1 with 1, 
2 with 2, etc., by solid lines. Also connect the points 
in the line of the top with those of the next lower 
number in the base, as 2 with 1, 3 with 2, etc., as in- 
dicated by the dotted lines. These solid and dotted 
lines form the bases of the two sets of triangles shown 
in the diagrams at the right, from which the correct 
measurements across the pattern are to be obtained. 

To construct these diagrams extend A 2 E 2 of tne 
side indefinitely to the right, as shown. At any con- 
venient points, as J and M, drop the perpendiculars J K 
and M N. From the points established in the quarter 
circle B 1 F 1 carry lines horizontally to the right, cut- 



Pattern Problems. 



379 



ting the two perpendiculars, as shown by the small 
figures above K and N. From J, upon J H, set off the 
space J 1, equal to the line 1 1 of the plan or top view, 
and from the point 1 thus established draw the hypoth- 
enuse of" the triangle, terminating in the point 1 of the 
line J K. In like manner set off from J, upon J H, the 
length 2 2 of the top view, also 3 3, 4 i and 5 5, and 
from the points thus established draw lines to points 
of corresponding designation in the line J K, all as 
shown. By this means triangles have been constructed 
the hypothenuses of which represent measurements on 
the surface of the finished article, taken on lines cor- 
responding to the solid lines of the top view. 

In like manner construct the second diagram of 
triangles shown at the extreme right, setting off from 




Fig. 661.— Pattern of Head Piece Shown in Fig. 660. 

M distances equal to the length of the dotted diagonal 
lines in the top view, and connecting the points thus 
established with points of next lower number in the 
line M N. Then the hypothenuses of this set of tri- 



angles will give the lengths corresponding to measure- 
ments on the dotted lines of the plan or top view. 

Having now obtained all the necessary measure- 
ments, the pattern may be developed as shown in Fig. 
661. The central portion of the pattern will corre- 
spond to A 1 C 1 B 1 of the end view, it being simply a 
flat triangular piece of metal. Therefore draw any 
horizontal line, as C B, equal in length to C B or C B 1 
of Fig. 660. Take the space 1 1 of the first diagrams 
of the triangles as radius, and from C and B, respect- 
ively, as centers, strike arcs which will intersect at A. 
From A as center, with 1 2 of the outer line of the 
plan as radius, describe a small arc. From B as cen- 
ter, with 1 2 of the second diagrams of triangles as 
radius, intersect the arc as shown, thus establishing the 
point 2 in the upper curve of the pattern. Then from 
B as center, with 1 2 of the arc B 1 F 1 of the end view 
as radius, describe another small arc, and from 2 of the 
upper edge of the pattern as center, with 2 2 of the 
first diagram of triangles as radius, intersect it as shown, 
thus establishing point 2 in the lower line of the pat- 
tern. Proceed in this way, using alternately the 
stretchout of the top of the tub, as indicated by the 
plan view, with the hypothenuses of the second dia- 
gram of triangles to establish the points in the upper 
curve of the pattern, and the stretchout of the quarter 
circle shown in the end view with the hypothenuses 
of the first diagram of triangles to establish the points 
in the lower line of the pattern, until the points 6 and 
5, or E and F, are reached. Connect E and F by a 
straight line and through the points from A to E and 
from B to F trace lines, as shown ; then A E F B will 
be the pattern for one of the corners of the head, a du- 
plicate of which may be reversed and transferred to the 
other side of the pattern, as shown by A D G C, thus 
completing the entire pattern of the head in one piece. 



PROBLEM 203. 

The Pattern for a Boss to Fit Around a Faucet. 



In Fig. 662 are shown two views of a boss such 
as is used for fastening a faucet into the side of a large 
can ; the curvature of the body of the can being rep- 
resented by the line A D B. For convenience in 
demonstration, what would be properly considered the 
front view of the article is here called the top view, 



the other view being considered as the side. Let H L 
and K 1ST represent its desired length and width of 
base or part to fit against the body of the can, and 
PRO the circle of the top to fit around the neck of 
the faucet. Also let D E be its required projection 
from the can. Through E draw Y Z parallel to H L, 



380 



Tae New Metal Worker Pattern Booh. 



the long diameter of the base. From P and drop 
lines at right angles to H L, cutting Y Z in the points 
Y and Z, also from H and L drop lines cutting A D B, 
and connect the points thus obtained with Y and Z, 
as shown, thus completing the side view. 

Commence by dividing one-quarter of the plan of 
the base K H into any convenient number of spaces, 
as shown by points 1, 2, 3, etc. For greater accuracy 
these spaces may be made shorter as they approach 
the ends of the base, where the line has more curve 
than near the middle. Having established the points 
0, 1, 2, 3, etc., in K H, draw a line from each of 
them to the center of the plan M. By this means the 
quarter of the circle representing the top of the article, 



measurements, as shown at the right of C B, the 
hypothenuses of which will represent the real distances 
between the required points. 

Therefore from the points established in H K 
drop lines vertically cutting the section line* ADB, 
as indicated, then carry lines from the points on A D 
horizontally till they cut the line C E and continue 
them indefinitely to the right. The points at which 
these lines cross the center line E C will represent the 
bights of the several triangles. On these horizontal 
lines, measuring from the center line C E, which is 
assumed as the common perpendicular for all the tri- 
angles, set off the bases of the several triangles, trans- 
ferring the distances from the plan. From the points 




Fig. 662.— Top and Side View of Boss, Showing System of Triangulation. 



and shown in the diagram by P R, will be divided in the 
same manner or proportionately to the plan of the 
base, all as shown by points l 1 , 2 1 , 3', etc. It will be 
seen that these lines divide the surface of the boss 
into a number of four-sided figures, each of which 
must now be redivided diagonally so as to form trian- 
gles. Therefore connect with 1', 1 with 2 1 , etc., 
by means of dotted lines, as shown. These solid 
and dotted lines drawn across the top view represent 
the horizontal distances between the points given, 
while the vertical distances between the same can be 
measured on lines parallel to C E ; hence it will be 
necessary to construct a series of triangles from these 



thus established draw lines to E, which will give the 
hypothenuses of the several triangles. For example, 
on the line drawn from the point 5 2 , in A D B, meas- 
uring from C, set off a distance equal to 5 1 5 and 
also a distance equal to 6' 5 in the top view. The 
difference between these two is so small as to be im- 
perceptible in a drawing to so small a scale as this. 
In like manner, on the line drawn from 4 3 set off a 
distance equal to the length of the diagonal lines 
i 5 1 and 4 4 1 in the top view, and in the same manner 
on the line drawn through 3 2 set off the distance equal 
to 3 -i 1 in the top view and also 3 3\ Then, as before 
remarked, lines drawn from the points thus established 



Pattern Problems. 



381 



in the horizontal lines toward E will be the hypoth- 
enuses of the several triangles corresponding to 
sections represented by the diagonal lines in the top 
view. 

In view of the fact that the base of the boss is 
curved as shown by A D it will be noticed that the 
measurements from K to H in the top view do not 
represent the real distances, because the distance H M 
is less than the distance A D. In case extreme ac- 
curacy is required it will therefore be necessary to 
develop an extended section on the base line A D, 
which may be done as follows : Extend the line M H 



required width of the pattern on one end, as shown by 
Y 6' in Fig. 662. With these two points established 
proceed to obtain other points in both lines of the 
pattern by striking arcs with radii equal to the spaces 
established in the plan of both base and top of the 
article and to the hypothenuses of the triangles already 
described. Thus, from S as center, with radius equal 
to the distance 6 3 5 4 of the stretchout of the base, de- 
scribe a short arc, as shown at 5 in the pattern. Then 
from T as center, with radius equal to E 6' of the tri- 
angles, intersect it by a second arc, as shown. From 
T as center, with radius equal to 6 1 5' of the plan of 




T ' u 

Fig. 663.— Pattern for Boss. 



of the top view, as shown at the left, upon which place 
a correct stretchout of A D ; that is, make D 1 l 3 equal 
to D 1 J , I s 2 3 equal to V 2% etc., and through each of 
the points thus obtained draw measuring lines at right 
angles to D 1 M. Place the T-square parallel to H M, 
and, bringing it successively against the points in the 
line K H, drop lines into the measuring lines of cor- 
responding number, as shown by 4 , l 4 , 2 4 , etc. Then 
will the distances 0* l 4 , l 4 2 4 , etc., be the correct dis- 
tances to be used in developing the pattern instead of 
the distances 1, 1 2, etc. 

The pattern may now be developed as shown in 
Fig. 663. Lay off the line S T, in length equal to the 



the top of the article, describe a small arc, as shown, and 
from 5 of pattern as center, and radius equal to E 5' of 
the triangles, intersect it by another arc, thus deter- 
mining the second point in the top. Proceed in this 
manner, adding one triangle after another in the order in 
which they occur in the top view, using the spaces of 
the plan of top and of the stretchout of the bottom and 
the hypotheneuses of the triangles as above described. 
Lines traced through the points thus obtained, as 
shown from S to N and from T to M, will give the pat- 
tern of one-quarter. This can be duplicated as often 
as is necessary to make the entire pattern in one piece, 
or to produce it in halves, as shown. 



PROBLEM 204. 



Patterns for a Ship Ventilator Having; a Round Base and an Elliptical Mouth. 



In Fig. 664 are presented the front and side ele- 
vations of a ship ventilator of a style in common use. 
A 1 B : shows the section or plan of its lower piece A E 
F B, as well as of the pipe to which it is joined, while 
EOSP is the shape of its mouth, or a section upon 



the line C D. The curves E C and F D connecting 
the two ends of the ventilator and forming the general 
outlines of the same may be drawn at the discretion of 
the designer. As the ventilator is constructed after 
the manner of an elbow, it may be divided into as 



382 



Tlie New Metal Worker Pattern Booh. 



many sections or pieces as desired. Therefore divide 
the curved lines E and F D into the same number 
of spaces, and connect opposite points by straight 
lines, as shown hj G II, K L and M 1ST. These lines 
should be so drawn as to produce a general equality in 
the appearance of the different pieces without refer- 
ence to equality in the spaces in either outline. 

The next step is to establish a profile or section 
upon each one of these lines. These profiles can be 
drawn arbitrarily, but each should be so proportioned 
that the series will form a gradual transition from the 
circle A 1 B' to the ellipse E S P. All the profiles 
will, therefore, be elliptical, those 
nearer the mouth being more elon- 
gated than those nearer the base or 
neck. Since the lower piece is 
cylindrical and is cut obliquely by 
E F, the section at E F must neces- 
sarily be a true ellipse and can be 
developed by a method frequently 
explained in connection with various 
problems in the first section of this 
chapter, and as also explained in 
Geometrical Problem 68 on page 
61. Of the remaining sections, their 
major axes are, of course, equal to 
the lengths of the lines G II, K L 
and M N, and their minor axes may 
be determined by any method most 
convenient, or in the following man- 
ner: Draw R U and S V, repre- 
senting a front view of the curved 
lines passing through the points n, 
m, Jc, g and e of the side view. From 
the points g, Jc and m project lines 
horizontally across to the front view, 
cutting the lines R U and S V and the center line T. 
Then f 1, do and b c will be respectively one-half 
the minor axes of the sections above referred to. With 
the major and minor axes of the several sections given, 
they may be drawn by any method producing a true 
ellipse, or in case the mouth has been drawn by means 
of arcs of circles the other sections may be drawn in 
the same manner. 

Each of the several pieces of which the ventilator 
is composed (except the lower piece) becomes, as will 
be seen, a transition piece between two elliptical curves 
not lying in the same plane, and in that respect is the 
same as the form shown in Problem 191. The pat- 
tern for each piece must, therefore, be obtained at a 



separate operation, that for the piece MNDC only 
being given. To avoid confusion of lines a duplicate 
of it is transferred to the opposite side of the front 
elevation, as shown by W Y Z X. Drop points from 
Y and Z perpendicular to the center line T of the 
elevation, thus locating the points M ! and N\ Make 
the distance tf c equal to b c. Then draw the ellipse 
Mr V W, which will be a front view of the sec- 
tion M 1ST of the side elevation. On a line parallel 
with Y Z construct the section M 1 Y N 1 , as follows : 
Let M' N 1 be equal to and opposite Y Z. Let the 
distance c' b' be equal to the distance c b of the see- 




ing!. 664. — Elevations and Sections of a Ship Ventilator. 



tion. With these points determined, draw through 
them the semi-ellipse M 1 5 1 N 1 . Divide the sec- 
tions M 1 S 1 N 1 and O S P into the same number of 
equal parts, as indicated by the small figures in the 
engraving. Drop the points 1, 2, 3, 4, etc., on to and 
perpendicular to the line Y Z ; thence carry them per- 
pendicular to the center line P of the front eleva- 
tion, cutting the section M" V W in the points l 2 , 2 3 , 
3 2 , etc., thus dividing it into the same number of 
spaces as were given to the original section M 1 b 1 N 1 . 
Next connect the points of like numbers in the two 
sections of the front elevation by solid lines, thus: 
Connect 2 1 with 2% 3' with 3 s , 4 1 with 4 2 , etc. ; also 
connect the points 2 1 with V, 3 1 with 2% 4 1 with 3\ 



Pattern Problems. 



383 



etc., by dotted lines, all as shown in the engraving. 
These lines represent the bases of right angled tri- 
angles, whose altitudes may be measured on the hori- 
zontal lines cutting the lines W X and Y Z. 

The next step, therefore, is to construct diagrams 
of these triangles, as shown at A and B of Fig. 665. 
Draw any two horizontal lines as bases of the triangles, 
and erect the perpendiculars E C and F D. On both 
E C and F D set off the various bights of the tri- 
angles, measured as above stated and as indicated by 
the points 1, 2, 3, 4, etc. Next set off the length of 
the. bases of the triangles as follows: In diagram A, 
let C 1 equal the distance l 1 1~ of Fig. 664; make C 2 
equal to 2 1 2 2 and C 3 equal to 3' o", etc. Connect 
the points in the vertical line with the points in the 
horizontal line of the same number, thus obtaining 





Fig. 665.— Diagrams of Triangles. 

hypothenuses of the triangles, or the true distance 
between the points l 1 1% 2 1 2 2 , etc., of the elevation. 
In diagram B, let the distances D 2, D 3, D 4, etc., 
represent the distances V 2 1 , 2 a 3 1 , etc., of the eleva- 
tion. Having located these points, connect 1 in the 
vertical line with 2 in the base ; also 2 in the vertical 
line with 3 in the base, and proceed in this manner 
for the other points. This will give the hypothenuses 
of the triangles, whose bases are l 2 2 1 , 2"- 3 1 , etc., in 
the elevation. 

Having thus obtained the dimensions of the vari- 
ous triangles composing the envelope of the first section 
of the ventilator, proceed to develop the pattern for 
it, as shown in Fig. 666. On any straight line, as C M, 
set off a distance equal to 1 1 in diagram A. From C 
as center, with radius equal to l 1 2 1 of the elevation, 



Fig. 664, draw an arc, which cut by another arc drawn 
from M as center, with radius equal to 1 2 of diagram 
B, thus establishing the point 2. From 2 as center, 
with radius equal to 2 2, diagram A, draw an arc, 
which cut with another arc drawn from l 1 as center, 
with radius equal to 1 2 of the elevation, thus estab- 
lishing the point 2\ Proceed in this manner, next 
locating the point 3, then the point 3 1 ; next the point 
4, and then 4', etc. It will be noticed that, after 




C M 

Fig. 666.— Pattern of First Section of Ship Ventilator. 

passing points 6 and 6 1 , 7 1 is obtained before 7. This 
is for the sake of accuracy, as it will be seen by in- 
spection of the elevation that the distance V 6 1 is less, 
and therefore more accurately measured in the eleva- 
tion, than the distance from 6 2 to 7\ Having thus 
located the points 1, 2, 3, etc., I 1 , 2', 3 1 , etc., trace 
the lines C D and N M, and connect D with 1ST, as in- 
dicated in Fig. 666. Then DNMC will be the 
pattern for one-half the section M N D C of the eleva- 
tion. 

The pattern of the section E A B F will be the 
same as that for the corresponding piece in an ordinary 
elbow, and, therefore, need not be specially explained 
here. 



384 



Tlie New Metal Worker Pattern Book. 

PROBLEM 205. 



Patterns tor the Junction of a Large Pipe with the Elbows of Two Smaller Pipes of the Same Diameter. 



The elbows of the smaller pipes in the problem 
here presented are such as would, if each were com- 
pleted independently of the other, form six-piece 
elbows. The junction between the two elbows occurs 
between the fifth pieces, which pieces unite to form 
the transition from the smaller diameters of the elbows 
to the diameter of the larger pipe, or sixth piece. A 
pictorial representation of the finished work is. shown 
in Fig. 667, in which, however, the upper section, or 
larger pipe, is omitted to more fully show the shape 
and junction of the transition pieces. A front view or 
elevation of the various parts is shown in Fig. 668. 
The side view given in Fig. 669 shows more fully the 
amount of lateral flare of the transition piece necessary 
to form a union between the varying diameters of the 
larger and smaller pipes. 

The drawing of that portion of the elbows in the 
smaller pipes from the horizontal parts up to the line 
a h t in Fig. 668 is exactly the same as that 
employed in drawing a six-piece elbow. The 
piece A G h a, occupying the place of what 
would otherwise be the fifth piece of the 
elbow, becomes in this case an irregular 
shape, the lower end or opening, ah, of which 
is nearly circular while its upper end, A G, is 
a perfect semicircle. This piece unites with 
its mate GD(/i on the line G h, thus form- 
ing the complete circle at A D, a plan of 
which is shown immediately below the ele- 
vation. The relative proportion between 
the diameters of the larger and smaller pipes 
is such that the junction between the elbows is carried 
somewhat below the fifth pieces, mitering the fourth 
pieces for a short distance, as shown from h to L. The 
method of cutting the lower parts of the elbow, how- 
ever, is the same as that employed in all elbow patterns 
where the pipe is of a uniform diameter throughout, 
numerous examples of which are given in Section 1 of 
this chapter, to which the reader is referred. 

As the section or profile of all the parts forming 
the elbow is a perfect circle when taken at right angles 
to the sides of the pipe, as at Q F or M N", it will be 
seen that a section on the line a h will be somewhat 
elliptical ; it will therefore be necessary to obtain a 
correct drawing of this section from which to obtain 
the stretchout of the lower end of the piece A G h a, 



with which it joins, and also a drawing of it as it will 
appear in plan. Therefore between two parallel lines 
drawn from M and N at right angles to M N construct 
a profile or section, as shown below at the left, which 
divide into any convenient number of equal spaces, as 
shown by the small letters «, b, c, etc. From each of 
these points carry lines back to M N at right angles to 
the same, and continue them in either direction till 
they cut the miter line a n of the elevation, as shown 
by the small letters, and the center line a n of the 
section. From the points in a n of the elevation draw 
lines at right angles to the same indefinitely, as shown 
above the elevation, across which at any convenient 
point draw a line, as B 1 C, at right angles to them. 
From B' C set off on the lines last drawn distances 
equal to the distances from the circumference to the 
diameter on corresponding lines in the section below, 
all as shown by a', h", c 2 , etc. A line traced through 




Fig. 667. — Perspective View of the Junction of a Large Pipe with 
tne Elbows of Two Smaller Pipes. 



these points will be the correct section on the miter 
line a n. It will be noticed that the section has not 
been carried further than the point Ji ', the balance of 
the curve not being required by reason of its intersec- 
tion with the corresponding piece in the other elbow. 

Below the elevation and in line with the same, as 
shown by the center line G T, is drawn the plan of the 
larger pipe ABCD. It will be necessary to add to 
this the plan of the curve on the line a h of the eleva- 
tion, in order that the horizontal distances between the 
points assumed in the two curves may be accurately 
measured. Therefore from the points on the miter 
line a h drop lines vertically through the plan, cutting 
the transverse center line X Y. From X Y set off 
distances on these several lines equal to the distances 



Pattern Problems. 



386 



of corresponding points from the line anoi the original 
section, as shown by a, b', etc., from X to S. A line 
traced through these points will give the correct posi- 
tion of the intersection of the smaller pipe as seen from 
above. This entire line is shown in the plan, although 
the part from S to Z will not be required, for the reason 
given above. An inspection of the plan will show that 
the side of the plan from V T to the right would be an 
exact duplicate of the left side if it were completed, 
and that therefore the plan consists of four symmetrical 
quarters, one of which, X E T, is completely shown 
in the plan. Hence the pattern for this quarter 
will suffice by duplication for the entire transition 
piece. 

Divide the quarter of the plan of the larger pipe 
P T, adjacent to the curve X S, into the same number 
of equal spaces as are found in 
the inner curve from X to S, 
as shown by the small figures 
1, 2, 3, etc. Connect corre- 
sponding points in the two lines 
as shown by the solid lines h' 
8, g' 7, /' 6, etc. Next sub- 
divide the four-sided figures 
thus obtained by their shortest 
diagonal, as shown by the dot- 
ted lines g' 8,/' 7, etc. These 
solid and dotted lines across 
the plan represent the bases of 
a series of right angled triangles 
whose altitudes can easily be 
obtained from the elevation, 
and whose hj^pothenuses when 
obtained will give correct dis- 
tances across the finished piece 
between points connected on 
the plan. These lines have 
also been drawn across the ele- 
vation from corresponding points in the same for 
illustrative purposes, but such an operation is not neces- 
sary to obtain the pattern. Neither is the side view 
shown in Fig. 669 necessary to the work, but is 
here introduced merely to assist the student in 
forming a more perfect conception of the operations 
described. From the points a, b, c, etc. on the miter 
line a h of the elevation carry lines horizontally across, 
cutting the vertical line Gr L, as shown by the points 
from s to h. The distances of these points from Gr will 
then represent the vertical distances of corresponding 
points in X S of the plan from the plane of upper base 



of the transition piece shown by A D of the elevation 
and V P T of the plan. 

To obtain the hypothenuses of the various tri- 
angles above alluded to, or in other words, the true 




Fig. 668.— Front Elevation, Flan and Sections, Showing Method of 
Triangulation. 

lengths of the lines dividing the surface, as shown in 
the two elevations and plan, it will be necessary to 
construct a series of diagrams, as shown in Fig. 670. 
Therefore draw any two lines, as A 8 and h h', at right 



386 



The New Metal Worker Pattern Book. 



angles to each other ; make h 8 equal to h' 8 of the 
plan, Fig. 668, and h h' equal to h 8 of the elevation, 
and draw h' 8. Next draw any two lines, as g 8 and g g', 
at right angles to each other, making g 8 equal to the 




Fig. 669.— Side Elevation. 

dotted line g' 8 of the plan and g 7 equal to the solid 
line g' 7 of the plan. Make g g' equal to the distance 
of point g from the line A D as measured by its cor- 
responding point on the line L G. Draw g' 8 and g 7. 
So continue till all the triangles have been constructed. 
Then the solid hypothenuses will represent the true 
distances across the pattern indicated by the solid lines 
of the plan and elevations, and the dotted hypothenuses 
the true distances on corresponding dotted lines in 



make equal to the line 1 a' of the diagram of triangles, 
Fig. 670. From 1 as a center, with a radius equal to 
1 2 of the plan, describe a small arc, which intersect 
with another small arc drawn from a as center, with a 
radius equal to a 2 of the diagram of triangles, thus 
locating the point 2 of the pattern. From point 2 as 
a center, with a radius equal to V 2 of the diagram of 
triangles, describe a small arc, which intersect with an- 




Fig. 671. — Pattern, for One-Quarter of Connecting Piece. 



other small arc struck from a of the pattern as center, 
with a radius equal to a 2 b 2 of the section on line a h 
of elevation shown above, thus establishing the posi- 
tion of the point b of the pattern. Proceed in this 
manner, using the spaces in P T in the plan of the 
larger pipe to form the upper edge of the pattern and 
the spaces from the section B 1 C to form the lower 
edge of the pattern, measuring the distances between 
the same by the alternate use of the solid and dotted 
hypothenuses of corresponding number and letter taken 
from the diagram of triangles in Fig. 670. A line 
traced through the two series of points and a straight 
line from S to h will complete the pattern for one- 




Fig. 670. — Diagram of Triangles. 



those views. In describing the pattern, work can be 
begun at either end of the pattern most convenient. 
Draw any straight line, as 1 a of Fig. 671, which 



quarter of the transition piece required. The remain- 
ing three-quarters can be obtained by any means of 
duplication most convenient. 



Pattern Problems. 3S7 

PROBLEM 206. 

The Patterns for a Right Angle, Two-Piece Elbow, One End of Which is Round and the Other Elliptical. 



In Fig. 672, let A G C B II D represent the eleva- 
tion of elbow, A F D the half profile of elliptical 
end and CEB the half profile of round end. The 
first step will be to establish a section on the miter 
line G H. Since the width at A D, one-half of which 
is shown by K F, is greater than J E, one-half the 
width at C B, it is proper that the width at L should 
be a medium between the two. Therefore from K, 
on K F, set off the distance J E, as indicated by K to. 
Bisect F m in n, and take K n as the width at L. The 
section at G H will then be an ellipse, of which G~ H 
is the major axis and K n one-half of the minor axis. 




Fig. 672. — Elevation of Elbow, with Half Profiles of the Two Ends. 

In Fig. 673, AGHD is a duplicate of the part 
bearing the same letters in Fig. 672. Against A D 
is placed a half profile, A F D, of the elliptical end, 
while against G H is placed one-half of the elliptical 
section, constructed as above described and as shown 
by G L H. Divide G L H into any convenient num- 
ber of equal parts, and from the points thus obtained 
drop perpendiculars cutting G H, as shown. Also 
divide A F D into the same number of parts, and from 
the points thus obtained drop perpendiculars cutting 
A D. Connect the points in A D with those in G H, 
as indicated by the solid and dotted lines. 

The next step is to construct sections on each of 



the solid and dotted lines drawn across the elevation 
by means of which to obtain the true distances between 
the points in A D and those in G H as though meas- 
ured upon the finished article. In Fig. 674 is shown 
a diagram containing sections upon the solid lines, 
which is constructed in the followina; manner : Draw 
any two lines, as M N* and M P, at right angles to each 
other. Upon M 1ST set off the hights of the several 
points in the profile A F D; thus make M 13, M 12, 
M 11, etc., respectively equal to k 13, j 12, h 11, etc., 
of Fig. 673. Upon M P set off from M the lengths 




Fig. 67S. — Elevation of Lower Piece of Elbow, Shoioing Method 
of Triangulation. 

of the several solid lines of the elevation ; thus make 
M a, M b, etc., respectively equal to / a, g b, etc., of 
Fig. 673, and at the points a, b, c, etc., thus obtained, 
erect perpendiculars, each equal in hight to the hight 
of the corresponding point in the profile G L H of 
Fig. 673 from the line G H. Thus make a 2, b 3, 



etc., of the diagram respectively equal to 



5 3, 



etc., of Fig. 673, and from the points 2, 3, 4, etc., 
thus obtained, draw solid lines to points 9, 10, 11, 
etc., in the lino M N, all as shown. Then the dis- 
tances 9 2, 10 3, 11 4, etc., will be the true lengths 
represented by corresponding solid lines drawn across 
the elevation. 



388 



Tlie New Metal Worker Pattern Book. 



The true distances represented by the dotted lines 
drawn across the elevation are obtained in the same 
manner by means of the diagram shown in Fig. 675. 
E S is drawn at right angles to R T and upon it are 
set off the hights of the points in A F D the same as 
in M N of Fig. 674. Upon R T set off from R the 
lengths of the several dotted lines drawn across the 



11 






t 
! 


: 


J 


io, ix; 


2 






6 


8,13 












M 













a b c d e 
Fig. 674.— Diagram of Sections Upon Solid Lines of Fig. 673. 

elevation, as shown by corresponding letters, and from 
the points thus obtained erect perpendiculars also as 
in Fig. 674. Finally connect by dotted lines such 
points as correspond with those connected by dotted 
lines in the elevation. Thus from 9 in R S draw a 
line to point 1 in the base line, corresponding to the 
line / 1 of the elevation, Fig. 673. Lines from 10 to 
2 and from 11 to 3 of the diagram will correspond 
respectively to g a and h b of Fig. 673. 

To develop the pattern from the dimensions now 
obtained proceed as follows : At any convenient place 



from H of pattern as center, describe an arc, which 
cut with another arc struck from point 9 of pattern as 
center, and 9 2 of Fig. 674 as radius, thus establish- 
ing point 2 of pattern. With point 2 of pattern as 
center, and 2 10 of Fig. 675 as radius, describe an 
arc, which intersect with another arc struck from point 
9 of pattern as center, and 9 10 of profile as radius, 



u 

10,12 



0,13 






--1 5 



R lab c d e 

Fig. 675. — Diagram of Sections Upon Dotted Lines of Fig. 67S. 

thus establishing point 10 of pattern. With point 10 
of pattern as center, and 1 3 of Fig. 674 as radius, 
describe a small arc, which intersect with one struck 
from point 2 of pattern as center, and 2 3 of G L H as 
radius, thus establishing point 3 of pattern. Continue 
this process, locating in turn the remaining points in 
pattern, as shown. Lines drawn through the points 
thus obtained, as indicated by G H and D A, will be 
one-half of the required pattern. The other half of 
the pattern can be obtained in a similar manner, 
or by tracing and transferring. The pattern for the 




li 13 12 11 .10 

A 

Fig. 676. — Pattern for Lower Piece of Elbow. 



draw the straight line H D in Fig. 676, in length equal 
to H D of Fig. 673. From H of pattern as center, 
with radius 1 9 of Fig. 675, describe an arc, which 
intersect by a second arc struck from D as center, with 
radius D 9 of profile, thus establishing the point 9 of 
pattern. Then with radius H 2 of GLH, Fig. 673, 



other part of elbow, as shown in Fig. 672 by Gr C B H, 
can be obtained by the same method. The shape 
G L H of Fig. 673 is to be drawn to the left of the 
miter line G H, and the operation continued, using 
the same process, as shown in Figs. 674, 675 and 
676. 



Pattern Problems. 



389 



PROBLEM 207. 

The Pattern for a Y Consisting of Two Tapering: Pipes Joining- a Larg-er Pipe at an Angle. 



In Fig. 677, BCDE represents the elevation of a 
portion of the larger pipe and 0' K D' L its profile. 
This pipe is cut off square at its lower end, with which 
the branches of the Y are to be joined. ABOHJ 
and G H E F are the elevations of the two similar 
branches joining each other from H to 0, and the larger 




Fig. 677.— Elevation, and Profiles of Y with Tapering Branches. 

pipe on the line BE. A' N J' M is the profile of one 
of the tapering branches at its smaller end. 

Since the article consists of two symmetrical halves 
when divided from end to end on the lines A' J' or C 
D' of the profiles, and since the two branches are alike, 
the pattern for one-half of one of the branches, as A B 
O H J, is all that is necessary. 

The dividing surface ABOHJ, lying as it were 
at the back of the half of the branch shown in eleva- 



tion by the same letters, will then form a plane or base 
from which the nights or projection of all points in the 
surface of the branch piece can be measured. 

As the branch piece A B H J is an irregular 
tapering form, its surface must be divided into a series 
of measurable triangles before its pattern can be ob- 
tained. Therefore divide the half profile C L D' into 
any convenient number of equal parts — in the present 
instance six, as shown by the small letters/^ hj k — and 
from these jDoints drop lines parallel with C B, cutting 
the line B E, as shown. In a similar manner divide 
the half profile A' N" J' into the same number of equal 




Fig. 67S. — Elevation of One Branch of Y, Showing Method of 
Triangulation. 

parts as was C L D', as shown by the small letters a 
b c d e. From the points thus obtained carry lines par- 
allel with J' J, cutting A J. Connect the points in 
A J with those in B E, as shown. 

To avoid a confusion of lines the subsequent oper- 
ations are shown in Fig. 678, in which A B H J is a 
duplicate of the piece bearing the same letters in Fig. 
677. The profiles B L E and A 1ST J are also dupli- 
cates of those shown in Fig. 677 and are for conven- 
ience here placed adjacent to the lines which they rep- 
resent. B L of the upper profile then represents a 
section on the line B 0, and A 1ST J that upon the line 
A J, but the section on the line H, the miter be- 
tween the two branches, is as yet unknown- To obtain 
this it will be necessary to first obtain sections upoa 



390 



Tlie New Metal Worker Pattern Booh. 



the various lines drawn across the elevation from B E 

to A J in Fig. 678, or in other words, diagrams upon 

which the true lengths of those lines can be measured. 

In the diagram of sections shown in Fig. 679 







S—— — 


9 


J, 


1 ' f. 


1 




a 




1 
1 

1 






Fig. 680. 
Diagrams of Sections Upon Solid Lines of the Elevation. 

S T represents the dividing surface or base plane al- 
luded to above and is made equal in length to 2 2' of 
Fig. 67S. At either extremity of this line erect the 
perpendiculars Si and T/, as shown. Make T/ equal 
in bight to 2' /of profile B L E, and upon S b set off 
from S the hight S a, equal to 2 a of the profile A N J, 
and draw the line a f. On S T, measuring from T, 
set off the distance 2' 2", and erect the perpendicular 
2" f", cutting af at /". Then will a f represent the 
true distance between the points 2 and 2' in Fig. 678, 
and af" will represent the true distance from 2 to 2", 




Fig. 681. 



df 



u 



T 6' 5' 
V 



Fig. 682. 
Diagrams of Sections Upon Dotted Lines of the Elevation. 

while 2" /" will be the hight of the point 2". In a 
similar manner set off from S, on S T, a distance equal 
to 3 3' of Fig. 678 and erect the perpendicular 3' g, 
equal in length to 3' g of profile B L E. - Make S b equal 
to 3 b of profile ANJ and draw b g. From S set off 



on S T a distance equal to 3 3" of Fig. 678 and erect 
the perpendicular 3" g", cutting b g at g" . Then will 
b g be equal to the true distance between 3 and 3' of 
Fig. 678, b g" will be the true distance from 3 to 3" of 
Fig. 67S and 3" g" will be the hight of the point 3". To 
construct the section on the line H, at points 3" and 
2" draw 3" g' and 2"/' at right angles to O H, making 
them respectively equal 3" g" and 2" /" of Fig. 679. 
As the profile B L E is a semicircle the hight of point 
•i' — that is, i' h — is equal to E ; therefore through the 
points E, g', f and H draw the curve shown, which 
will be the true section on line H, from which the 
stretchout can be taken for that portion of the 
pattern. 

The sections on the remaining' lines (i 4', 5 5' and 
6 6') of the elevation are shown in Fig. 680 and are 
constructed in exactly the same manner as those shown 
in Fig. 679, giving c h, dj and e k as the true lengths 




Pattern of Tapering Branch. 



of those lines. Before the pattern can be developed 
the four-sided figures into which the surface of the 
branch pipe has been divided by the solid lines must 
be subdivided into triangular spaces, as shown by the 
dotted lines in the elevation. Sections upon these lines 
must also be constructed, in order that their true 
lengths can be obtained. These are shown in two 



groups in Figs. 681 and 682 and are constructed in a 
manner exactly similar to that described in connection 
with Fig. 679. They may be easily identified by cor- 
respondence between the figures on the base lines U V 
and W X and those of the elevation. 

To describe the pattern proceed as follows : Draw 
any line, as J H in Fig. 683, in length equal to J H of 
Fig. 678. With J of pattern as center, and J' a of 
smaller profile as radius, describe a small arc (a), which 



Pattern Problems. 



391 



cut with one struck from H of pattern as center, and 
1" a of Fig. 681 as radius, thus establishing the point 
a of pattern. With a of pattern as center, and a f" of 
Fig. 679 as radius, describe another small arc (/'), 
which intersect with one struck from H of pattern as 
center, and H/' of profile H E as radius, thus estab. 
lishing the point /of pattern. In a similar manner, a 
b of pattern is struck with a b of profile as radius ; /' b 
of pattern with/" b of Fig. 681 as radius; b g' of pat- 
tern with b g" of Fig. 679 as radius, and/' g' of pattern 
with/' g' of profile HE as radius; also, b c of pattern 
is struck with b c of profile as radius ; g' c of pattern 
with g" c of Fig. 681 as radius ; g' h of pattern with 
g E of profile as radius, and c h of pattern with c h of 



Fig. 680 as radius. Thus are the points established 
in O H J P of pattern. 

B P A of pattern corresponds with B P A 
of elevation and is obtained in the same manner. The 
points in B of pattern are derived from profile L B, 
as are the points in P A of pattern from N A of small 
profile. The lengths of solid lines in pattern are ob- 
tained from the diagram of sections in Fig. 680, as are 
those of the dotted lines from the diagram of sections 
in Fig. 682. Lines drawn through the points in B 
II J P A, Fig. 683, will be the half pattern for ABO 
H J of elevation. The other half of pattern, as shown 
by A J' H' 0' B, can be obtained in a similar manner 
or by duplication. 



PROBLEM 208. 



Pattern for a Three-Pronged Fork With Tapering Branches. 



In Fig. 684 is shown a pictorial representation of 
a fork, or crotch, consisting of three branches of equal 
size and taper; all uniting so as to form one round 
pipe. 

In the plan, Fig. 685, ABC represents the base 
of article or size of the large pipe and B D E C G one 
of the tapering branches. The other branches are 
partly shown in plan by A G C S T and AUYBS. 




Fig. 684- — Perspective View of Three-Pronged Fork with Tapering 
Branches. 

In the elevation the branch is shown by J K L M N 
and the half profile of small end by K E, L. 

An inspection of the engraving will show that the 
perimeter of the larger end of the branch must be di- 
vided into three parts, two of which form the joints or 
-connections with the branches on either side of it 



while the third part must form one -third of the base 
or circumference of the large pipe with which it is to 
be united. In the elevation P M represents the plane 
of the base or upper end of the round pipe of which 
A B C is the profile or plan, and J is assumed as 
the hight of the central point at which all the branches 
meet. From J of the elevation or G of the plan to 
either of the three points A, B or C any suitable curve 
may be chosen as the profile upon which to make a 
joint or miter between adjacent branches. As J is 
equal to G A or G C, a quarter circle is assumed as the 
most suitable curve ; therefore from as a center de- 
scribe the quarter circle P J of the elevation, corre- 
sponding with A G of the plan. In order to complete 
the elevation of the branch J K L M N, it will be 
necessary to obtain the elevation of the miter line G C. 
Therefore divide P J into any convenient number 
of equal parts, as shown by the small figures, and from 
the points thus obtained carry lines to the right paral- 
lel with P M. From G, on G C, set oil spaces equal 
to the distances from the points in P to the line 
O J, as shown, and from the points thus obtained 
in G C erect perpendiculars cutting lines of similar 
number drawn from P J. A line traced through these 
points of intersection, as shown by J N, will give the 
miter line in elevation corresponding with G C of the 
plan. Divide C H of the plan into the same number 



392 



Tlie New Metal Worker Pattern Book. 



of equal parts as P J of the elevation, and from the 
points thus obtained erect perpendiculars cutting N M. 
Divide K E L, the profile of the smaller end of the 
branch, into the same number of equal parts as the 
larger end — that is, as many as are found in J N M — 
and from the points of division drop lines perpen- 
dicular to K L, cutting the same. Connect points in 
K L with those in J N M by solid and 
dotted lines in the manner shown in 
the drawing. Upon all of these lines 
it will be necessary to construct sec- 
tions in order to obtain the true dis- 
tances as if measured upon the surface 
of the branch. As each of the branch 
pipes consists of symmetrical halves 
when divided by the line Gr F of the 
plan half sections only need be con- 
structed, all projections being measured 
from the dividing plane represented by 
G F in the plan and shown in eleva- 
tion by JKLM N. 

In Fig. 6SG are shown the sec- 
tions having for their bases the solid 
lines of the elevation, which are con- 
structed in the following manner : 
Upon any horizontal line, as P Q, set 
off from P the lengths of the several 
solid lines of the elevation, as indicated 
by the small figures corresponding with 
those in JN M. At P, which corre- 
sponds with all the points in K L of 
the elevation, erect a perpendicular, P 
H, upon which set off the bights of the 
points in K R L, as 2' 2, 3' 3, etc., 
shown by P 2, P 3, etc. At each of 
the points near Q erect a perpendicu- 
lar, which make equal in hight to the 
length of line drawn from the point of 
corresponding number in G C H of 
the plan to the line G H. Thus make 
9' 9, 10' 10, etc., equal to 9" a, 10" 
b, etc., of the plan. From the points 9, 
10, etc., draw solid lines to the points in H P, connect- 
ing points correspondingly connected by the solid lines 
of the elevation. The sections having for their bases 
the dotted lines of the elevation are shown in Fig. 687, 
and are constructed in exactly the same manner. 
Upon Y Z, set off from Y the lengths of the clotted 
lines of the elevation, numbering the points near Z 
to correspond with those in JNM of the elevation. 



The perpendiculars erected from these points are the 
same as those similarly located in Fig. 6S6, and the 
perpendicular X Y is a duplicate of H P of Fig. 6S6. 
From the points 9, 10, 12, etc., draw clotted lines to 
points in X Y, connecting points correspondingly con- 
nected b} r dotted lines of the elevation. 

To describe the pattern shown in Fig. 6SS pro- 



ELEVATION 
J 




Fig. (JS5. — Plan and Elevation of Three-Pronged Fork. 

ceed as follows : Draw any line, as J K, in length 
equal to J K of elevation, Fig. 685. With K of pat- 
tern as center, and K 2 of profile as radius, describe a 
small arc (2), which cut with one struck from J of 
pattern as center, and 8' 2 of Fig. 687 as radius, thus 
establishing point 2 of pattern. With point 2 of pat- 
tern as center, and 9 2 of Fig. 686 as radius, describe 
another small arc (9), which intersect with one struck 



Pattern Problems. 



393 



from J of pattern as center, and J 9' of elevation as 
radius, thus establishing the point 9 of pattern. Pro- 




Fig. 686. — Diagram of Sections Upon Solid Lines in J K L M N 
of Fig. 6S5 

ceed in this manner until the remaining points are 
located, all as clearly indicated by the solid and dotted 
lines in Fig. 688. By drawing lines through the 






13 



35 
26 



8'12')13' ' 

10 /u 

Fig. 6S7 .—Diagram of Sections Upon Dotted Lines in J K L M N 
of Fig. 6S5. 



points thus obtained the half pattern shown by K Q 
LMN J is the result. The other half, as shown by 



N li 




J s fr 



Fig. 688.— Pattern of Tapering Branch. 

K Q' 1/ M' 1ST J, can be obtained in a similar maDner, 
or by duplication. 



PROBLEM 209. 



The Pattern for an Offset to Join an Oblong Pipe With a Round One. 



In Fig. 6S9, B C F Gr represents the side elevation 
of the offset, ABGHa portion of a round pipe join- 
ing it below, and C D E F a portion of the oblong 
pipe joining it above. In the plan immediately below, 
JKLM shows the plan of the round pipe and 1ST P 
QES that of the oblong pipe, while the distance L T 
shows the amount of the offset. 

The piece forming the offset is similar in shape 
to that shown in Problem 189, the difference being 
that its bases B Gr and C F are neither horizontal nor 
parallel to each other and that sections on the lines of 
the bases are not given. Since the article required 
consists of symmetrical halves when divided on . the 
line J T of the plan, the plane surface ABCDEFGH 



lying as it were back of the half shown by the eleva- 
tion may, as in Problem 207, be regarded as a base 
from which to measure all bights, or projections, in 
obtaining the required profiles and sections necessary 
in developing the pattern. The first steps necessary 
will be to obtain true sections upon the lines C F 
and B Gr of the elevation. In Fig. 69 0, CDEF rep- 
resents a duplicate of the part bearing the same letters 
in the elevation. Upon D E as a base line construct 
a duplicate of the half section of oblong pipe NOPT 
of Fig. 689, as shown by D N" P E. 

Divide the semicircle 1ST P- into any convenient 
number of equal parts, as shown by the small figures. 
With the blade of the T-square placed at right angle? 



39i 



Tlie New Metal Worker Pattern Book. 



toDE, drop lines cutting C F. With the T-square 
placed at right angles to G F, and brought against the 
points in C F, draw lines, extending them indefinitely, 
as shown. Measuring in each instance from C F, set 
off on the lines just drawn the same length as similar 
lines in D K P E, and through the points thus ob- 




Fig. 6S9.—Plan and Elevation of Offset. 

tained trace a line, as shown by n Op. Then CFj) 
O n is the half shape of cut on line C F. In Fig. 691, 
A B G H is a duplicate of the elevation of the round 
pipe, below which is drawn a half profile of same, A 
M H. To obtain the shape of cut on line B G, divide 
the half profile A M H into the same number of parts 
as was N P, and, with the T- sc L uare placed parallel 



with A B, and brought successively against the points 
in A M H, carry lines cutting B G. With the T-square 
placed at right angles to B G, and brought against the 




Fig. 690. — Development of Section on Line C F of Elevation. 

points therein contained, erect perpendiculars, as shown. 
Measuring in each instance from B G, set off on the 
lines just drawn the same length as similar lines in A 




14 h 



Fig. 691. — Development of Section on Line B G of Elevation. 

M H, and through the points thus obtained trace a line, 
as shown by B on G. Then B in G is the half shape 
of cut on line B G. 

In order to avoid a confusion of lines a duplicate 



Pattern Problems. 



395 



of B C F G of the elevation is presented in Fig. 692, 
upon C F and B G of which, as base lines, are drawn 
duplicates of the sections obtained in Figs. 090 and 
691, all as shown. From points in n 0^9 drop lines at 
right angles to C F, cutting the same, and from points 
in B m Gr drop lines at right angles to B Gr, cutting it. 
Connect points in these lines in consecutive order by 
solid lines, as shown, and subdivide the four sided 
figures thus obtained by dotted lines representing their 
shorter diagonals. The surface of the offset or transi- 
tion piece is thus divided into a series of very tapering- 
triangles, the lengths of whose bases or shortest sides 




Fig. 692. — Middle Piece of Offset, Showing Method of Triangulation. 

are given in the two sections C nOjjF and B m G. 
In order to obtain the correct lengths of their longer 
sides two diagrams or series of sections must be con- 
structed for that purpose, which are shown in Figs. 
693 and 694. 

To obtain the various sections on the solid lines 
of the elevation proceed as follows : Draw the right 
angle UTW, Fig. 693. From V, on V U, set off 
the length of lines in B m Gr, Fig. 692. From V, on 
V W, set off the length of solid lines in B C F G, and 
from the points thus obtained erect perpendiculars, in 
length equal to lines of similar number in C n p F, 
Fig. 692. Thus make line W V of Fig. 693 equal to 



line C n or C 1, and draw VI', which gives the 
distance from point B to point n in Fig. 692 as if 
measured on the finished article. Connect the perpen- 
diculars drawn from V W with the points in V U, as 
shown, and corresponding with the figures in Fi°\ 692. 
Thus connect 1' and 8, 2' and 9, 3' and 10, etc. Then 




7 6 5 4 3 2 1V/' 
Fig. 693.— Diagram of Sections on Solid Lines of Fig. 692. 

will the lengths of the oblique lines in Fig. 693 be the 
true lengths of the solid lines crossing the elevation. 
The diagram of sections shown in Fig. 694 is con- 
structed in the same manner, using the dotted lines of 
the elevation as the basis of measurements. 

Draw the right angle X Y Z, and from Y set off 
on Y Z the length of lines in G n p F. From Y, on 
Y X, set off the length of dotted lines in B C F G, and 
from the points thus obtained erect perpendiculars, in 
length equal to lines of similar number in B m G, as 
indicated by the small figures. Connect the perpen- 
diculars drawn from X Y with the points in Y Z, as 
shown, and corresponding with the figures in B C F G. 
Thus connect 1 and 9', 2 and 10', 3 and 11', etc. 



Z 4 



^«=^=^~ 



n 

lo'r — 



'-», 



1112 j^ 

13| 
I 
14 



2-6 
1-7 



Fig. 694. — Diagram of Sections on Dotted Lines of Fig. 692. 

An inspection of the plan and elevation, Fig. 689, 
will show that the curved surface of the offset or transi- 
tion piece B C F G, which has been divided into tri- 
angles, is shown by J M L Q R S of the plan, and that 
this piece is connected with its mate or equivalent in 



396 



Trie Neiv Metal Worker Pattern Book. 



the opposite half of the article by a large plain trian- 
gular surface, S J 1ST, on the upper side, and by another, 
Q L P, on its lower side, which must be added to the 
pattern of the curved portion after it has been de- 
veloped. It will also be seen that V W 1' of Fig. 
693 is one-half of J N" S. Therefore to develop the 
pattern, first draw any line, as j x in Fig. 695. in length 
equal to B C of Fig. 689, or V W of Fig. 693. With 
j as center, and 1' V of Fig. 693 as radius, describe a 
small arc (near s), which intersect with another small 



intersect with another small arc struck from point 1 of 
pattern as center, and a radius equal to 1 2 of the pro- 
file C n p F of Fig. 692, thus establishing the position 
of point 2 of pattern. Proceed in this manner, using the 
dotted oblique lines in Fig. 694, the lengths of the 
spaces in B m G in Fig. 692, the lengths of the solid 
oblique lines in Fig. 693 and the lengths of the spaces 
in C n p F of Fig. 692 in the order named until the 
line 7 1-t is reached. Lines traced through the points 
of intersection from j to I and from s to t will give the 




Fig. 695.— Pattern of Offset. 



arc struck from x as center, and with a radius equal to 
W V of Fig. 693, thus duplicating the triangle V W 
1'. From s, or point 1, as center, with a radius equal 
to 1 9' of Fig. 694, describe a small arc (near 9), which 
intersect with another small arc struck from j or 8 of 
pattern as center, with a radius equal to 8 9 of the pro- 
file B m G, Fig. 692, thus establishing the position of 
point 9 of pattern. From 9 as a center, with a radius 
equal to 9 2' of Fig. 693, describe a small arc, which 



shape of the curved portion of the pattern. From 14 
of pattern as center, with a radius equal to G F of Fig. 

692, or V 7 of Fig. 693, describe a small arc, which 
intersect with another small arc struck from point 7 of 
pattern as center, and a radius equal to 7 7' of Fig. 

693, or T P of the plan. Draw 7 t and t I; then will 
I j x s g t he one-half the pattern required. The other 
half can be obtained by any means of duplication 
most convenient. 



PROBLEM 210. 

Pattern for an Offset to Join a Round Pipe with one of Elliptical Profile. 



This problem differs from the preceding one only 
in the shape of the pipe having the elongated profile, 
which profile in the preceding problem consists of two 



semicircles joined by a straight part, whereas in this 
case its curve is continuous throughout; its pattern 
therefore will consist throughout of a series of triangles 



Pattern P ruble 



ms. 



397 



having short bases instead of having; a large flat trian- 
gular surface uniting its curved portion as in the pre- 
vious case. 

In Fig. 696, DCB A represents the elevation of 
the offset, CFE B that of a portion of the round pipe 
with which it is required to connect at its upper end 




Fig. 696. — Elevation and Sections of Offset, Showing Method of 
Triangulation. 

and HDAG that of the elliptical pipe joining it be- 
low. MPNis the half profile of the round pipe and 
K J L that of the elliptical pipe. The plan or top 
view is not shown, and is not necessary to the work of 
obtaining the pattern. Since the profiles given neces- 
sarily represent sections on lines at right angles to the 
respective pipes, as at F E and H G-, it will first be 
necessary to derive from them sections on the joint or 



miter lines C B and D A, from which to obtain correct 
stretchouts of the two ends of the pattern of the offset 
piece. 

As the pattern required consists of symmet- 
rical halves, one-half only will be given, and one-half 
of the profiles only need be used. Therefore divide 
the half profile M P 1ST into any convenient number of 
equal spaces, as shown by the small figures, and from 
the points thus obtained draw lines at right angles to 
F E, cutting M 1ST and C B. To avoid confusion of 
lines a duplicate of C B is shown at the left by C 1 B 1 . 
From the points on C B 1 draw lines at right angles to 
it indefinitely, and upon each of these lines, measuring 
from C B', set off the lengths of tines of corresponding 
number in the profile M P N measured from M N. Thus 
make the distance of point 2' from C 1 B 1 equal to the 
distance of point 2 from line M N, the length of line 
3' equal to that of line 3, measuring from the same base 
lines as before, etc. A line traced through the points 
of intersection, as shown by C 1 B 1 , will be the correct 
section on the line C B of the elevation. The method 
of obtaining the section on the line D A, shown at D 1 
I A 1 , is exactly the same as that just described in con- 
nection with the round pipe, all as clearly shown in the 
lower part of the engraving. 

The next operation will consist of dividing the 
surface of the transition or offset piece into measurable 
triangles, making use of the spaces used in the profiles ; 
therefore connect points in C B with those of similar 
number in D A by solid lines, as 1 with 1', 2 with 2', 
etc., and connect points in C B with those of the next 
higher number in D A by dotted lines, as 1 with 2', 2 
with 3', 3 with 4', etc. The surface of the transition 
piece is thus divided into a series of triangles the 
lengths of w T hose bases or short sides are found in the 
two sections C 1 B 1 and D 1 I A 1 . 

As the bights of corresponding points in the two 
sections, measuring from their center or base lines, 
differ very materially, it will be necessary to construct 
two diagrams of sections from which the lengths of the 
various solid and dotted lines can be obtained. In 
Fig. 697 is shown a diagram of sections through A B 
C D taken on the solid lines drawn across the eleva- 
tion, in which the base line P Q represents the sur- 
face of a plane dividing the offset into symmetrical 
halves. At P erect a perpendicular, P R, upon which 
set off the hight of the points in the profile K JLor 
the section D 1 I A 1 , measuring upon the straight lines 
joining them with the base line K L, as shown by the 
small figures. From P, upon P Q, set off the lengths 



S9S 



Tlie New Metal Worker Pattern Booh. 



of the various solid lines drawn across the elevation, 
also shown by small figures, and at each of the points 
thus obtained erect a perpendicular, which make equal 
in hight to the distance of point of corresponding num- 
ber in profile M P N from M N, measuring on the per- 
pendicular line. Thus, make line 2 of Fig. 697 equal 




Fig. 697 .—Diagram of Sections on Solid Lines of Elevation. 

in hight to the distance from point 2 of profile to the 
line M N, line 3 equal to the length of line 3 of profile 
M P N. Now connect points of corresponding num- 
ber at the two ends of the diagram by straight lines, 
as shown, then will these oblique lines be the correct 
distances between points of corresponding numbers 
connected by the solid lines drawn in the elevation. 

The diagram in Fig. 69 S is constructed in an ex- 
actly similar manner. The distances S 1, S 2, S 3, 
etc., on the base line are in this case made equal to the 
lengths of the dotted lines of the elevation, and the 




Fig. 69S. — Diagram oj Sections on Dotted Lines of Elevation. 

perpendiculars erected at points 2, 3, etc., are the 
same as those used in the previous diagram. The per- 
pendicular S U is also an exact duplicate of P E in 
Fig. 697. In drawing the oblique dotted lines, point 
1 at the right end of the diagram is connected with 
that of the next higher number (2') on the line S U, 2 
at the right with 3' on the line S U, etc., all as shown. 



The oblique dotted lines will then be the correct dis- 
tances between points of corresponding numbers con- 
nected by the dotted lines in the elevation. 

To develop the pattern, first draw any straight 
line, as D C in Fig. 699, which make equal in length to 
DC of Fig. 696. From D as center, with a radius 
equal to 1 2 of the section D' I A 1 , strike a small arc, 
which intersect with another small are struck from G 
as center, with a radius equal to 2' 1 of Fig. 698, thus 
establishing the location of point 2' of pattern. From 
2' of pattern as center, with a radius equal to 2' 2 of 
Fig. 697, strike a small arc, which intersect with an- 



\0B 




Fig. 699.— Half Pattern of Offset Piece. 

other small arc struck from C of pattern as center, and 
a radius equal to 1 2 of section C B 1 , thus establish- 
ing the position of point 2 of pattern. So continue, 
using alternately the dotted and the solid oblique lines 
in Figs. 6.9S and 697 to measure the distances across 
the pattern, the spaces from the section D 1 I A 1 to 
form the stretchout of the lower end (D A) of pattern, 
and the spaces from the section C 1 B 1 to form the 
stretchout of the upper end (G B) of the pattern. Lines 
traced through the points of intersection, as from C to 
B and from D to A, will complete one-half the required 
pattern. 



Pattern Problems. 

PROBLEM 211. 
The Patterns for a Funnel Coal Hod. 



399 



In Fief. 700 are shown the drawings for a funnel 
coal hod of a style in general use. In preparing such 
a set of drawings it is necessary that care should be 
taken to have a correspondence of all the principal 



here constructed in two pieces, the front being in one 
piece joined together on the line B C of the elevation 
or B' 2 C 3 of the plan, and joined to the back piece on 
the line II D. As will be seen by an inspection of 




Fig. 700 — Plan, Elevation and Sections of a Funnel Coal Hod, Showing Method of Triangulation. 



parts in the two views, as shown by the dotted lines, 
leaving the final drawing of the curves to be more 
accurately performed as circumstances may require in 
subsequent parts of the work. The design is capable 
of any degree of modification so far as the proportions 
of its parts are concerned without in the least affect- 
ing the method of obtaining its patterns. Thus, 
hights, lengths, diameters or curves may be changed 
at the discretion of the designer. The coal hod is 



the elevation, the front piece consists of a flat tri- 
angular piece, H J D, joined to two irregular flaring 
pieces, A J II G and B J D C. On account of the 
taper or slant of the flat portion of the front piece, as 
shown by J 2 D = of the plan, the line D~ IF has been 
drawn somewhat obliquely from X, the center of the 
bottom, instead of at right angles to A 2 E\ 

The section at A B is assumed to be a perfect 
circle and should be drawn exactly opposite, as 



400 



The New Metal Worker Pattern Book. 



shown, its vertical centsr line A 1 B 1 being placed 
parallel to A B. Divide each quarter of this, as A 1 J' 
and J 1 B l , into any number of equal spaces, as shown 
by the small figures, and through the points thus ob- 




Q b i U u 

Fig. 701.— Diagram of Sections on Solid Lines in A J H G of Fig. 700. 

tained draw lines cutting A 1 B 1 and A B. From J, 
the middle point on A B, draw lines to D and to H. 
Also divide H' G 1 of the plan into the same number 
of equal spaces as A 1 J 1 , numbering the points to cor- 
respond. From the points thus obtained erect lines 
perpendicularly, cutting G H of the elevation. Con- 
nect points of like number on A J and G H, as 5 with. 
5, 6 with 6, etc., by solid lines, as shown; also, con- 
nect each point on A J with that of next higher 
number on G II by a dotted line, as 5 with 6, 6 with 
7, etc. These solid and clotted lines just drawn are 
the lines upon which measurements are to be taken in 
obtaining the pattern, and upon which sections must 



5 s 

6 

7 
8 
9 



,u 



7 9S 7 6 

Fig. 70S.— Diagram of Sections on Dotted Lines in A J H G of Fig. 700. 

be constructed before their true lengths can be ob- 
tained. 

In Fig. 701 are shown the sections having the 
solid lines in A J II G as their bases, which are con- 
structed in the following manner : Draw any right 
angle, as P Q R. Upon P Q set off the hights of the 
several points in the section A' J' from the line A' B 1 , 
as measured upon the straight lines joining them with 
A 1 B 1 ; thus make Q 5 and Q 6 equal to the distance 
of points 5 and 6 from the line A 1 B 1 . From Q on 
Q B, measuring from Q, set off the lengths of the 
several solid lines in A J H P, as indicated by the 
small figures, and from the points thus obtained erect 
perpendiculars equal in hight to the length of lines 
drawn from points of corresponding number in G 1 H 1 
of the plan to the line G 1 X ; thus make the perpen- 



diculars at points 5', 6', etc., equal to the length of 
the lines drawn from points 5 and 6 in G 1 H 1 to G' X. 
Connect the points thus obtained with points of cor- 
responding number in P Q. The oblique lines thus 
obtained will be the true distances represented by 
lines of corresponding number in the elevation. The 
diagram in Fig. 702 shows the sections upon the dotted 
lines in A J H G and is constructed in the same man- 
ner. Upon T U, measuring from T, are set off the 
lengths of the several dotted lines. S T is the same 
as P Q of Fig. 701, and the perpendiculars at U are 
equal to' those of corresponding number in Fig. 701. 
Points in S T are then connected with the perpen- 
diculars of next higher number by dotted lines, which 




L 23 4 s 

Fig. 70S.— Diagram of Sections on Solid Lines in J B O D of Fig. 700. 

give the true lengths represented by the dotted lines of 
the elevation. 

That portion of the front piece shown by J B C D 
of the elevation must be triangulated in exactly the 
same manner as the portion just described, and sec- 
tions constructed upon the several solid and dotted 
lines there drawn, as shown in Figs. 703 and 701. 
However, as no outline is given in either the plan or 
the elevation from which a correct stretchout of C D 
can be obtained, a section must be constructed for that 
purpose, which can be done in the following manner : 
First draw C M 1 as the vertical center line of a rear 



,Y 



- W IS 3 4 

Fig. 704.— Diagram of Sections on Dotted Lines in J B C D of Fig. 700. 

elevation. From points C and D project lines hori- 
zontally to the right, cutting C M 1 at C and M. Upon 
D M, measuring from M, set off half the width of the 
front piece at D" of the plan; that is, make M D' 



Pattern Problems. 



401 



equal to M 2 D\ Any desirable curve may then be 
drawn from D 1 to C, representing the rear elevation of 
curve represented by D G of the side elevation. As 
the distance from C to D is much greater than C 1 M, 
an extended profile, as measured upon C D, must now 
be developed from which to obtain a correct stretch- 
out of that portion of the pattern. 

Therefore divide the curve C 1 D 1 into the same 
number of parts as the quarter circle B 1 J 1 , and from 
the points thus obtained cany lines horizontally to the 
left, cutting C D. Upon C M extended, as C a M 1 , set 
off spaces equal to those in C D, as shown, and through 




Fig. 70 J. — Half Pattern of Froyit Piece of Funnel Coal Sod. 



the points thus obtained draw lines to the left indef- 
initely. From the points in C D 1 drop lines verti- 
cally, cutting those just drawn, all as shown. A line 
traced through the points of intersection, as shown 
from D 2 to C 2 , will give the desired stretchout. In 
dividing the curve C D 1 into spaces it is advisable to 
make those nearest to D 1 less than those near the top 
of the curve in oixler to compensate for the increase in 
the spaces in C" M 1 as they approach the bottom ; 
thus obtaining a set of nearly equal spaces upon the 
final profile D 2 (J 2 , all of which will appear clear by an 
inspection of the drawing. 

As above stated, the diagrams of sections in Figs. 
703 and 704 are constructed in the same manner as 



those of Figs. 701 and 702. The bights in K L aud 
V AV are taken from J' B 1 of Fig. 700 and are the 
same as those in P Q and S T of Figs. 701 and 702. 
The distances upon L N and "W Y are those of the 
solid and dotted lines in J B C D of Fig. 700, and the 
bights of the perpendiculars near N and Y are equal 
to the lengths of the lines drawn from points of cor- 
responding number in the profile D 2 C 2 of Fig. 700 to 
the lines C 2 M 1 . 

To develop the pattern of the front piece, first 
draw an)- line, as A G in Fig. 705, equal in length to 
A G of Fig. 700. From G as a center, with a radius 
equal to the dotted lines 9 8 of Fig. 702, describe a 
short arc (near S), which intersect with another arc 
drawn from A as center, with a radius equal to 9 8 of 
the section A 1 J 1 B 1 of Fig. 700, thus establishing the 
position of point 8 in the upper line of the pattern. 
From 8 of the pattern as center, with a radius equal 
to 8 S of Fig. 701, describe a short arc (near 8'), which 
intersect with another arc drawn from G of the pat- 
tern as center, with a radius equal to 9 8 of the plan, 
Fig. 700, thus establishing the point S' in the lower 
line of the pattern. Continue in this manner, using 
the lengths of the oblique clotted lines in Fig. 702 in 
connection with the spaces in the section A' J 1 B' of 
Fig. 700 as radii to determine the points in the upper 
line of the pattern, or the side forming the mouth, and 
the lengths of the oblique solid lines of Fig. 701 in 
connection with the spaces in the plan of the bottom 
(G 1 H 1 ) as radii with which to determine the points in 
the lower line of the pattern or the side to fit against 
the bottom. 

Having reached the points 5 and 5', next add to 
the pattern the flat triangular surface shown by J II D 
of the elevation. From II (5') of the pattern as center, 
with a radius equal to 5 5 of Fig. 706, the Side of the 
last triangle in the pattern of the back piece, describe 
a short arc (near D), and intersect the same with 
another arc struck from J (5) of the pattern as center, 
with a radius equal to the oblique line 5 5 of Fig. 703, 
and draw II D and D J. Using D J of the jjattern as 
one side of the next triangle, take as radii the dis- 
tances 5 4 of Fig. 701 and 5 1 of the section D 2 C 2 of 
Fig. 700 to locate the position of point 1' of the pat- 
tern, as shown in Fig. 705. With 1 4 of Fig. 703, 
and 5 4 of the section B' J' of Fig. 700 as radii locate 
the point 4 of the pattern, as shown, and so continue 
until C D is reached. Lines traced through the points 
of intersection from B to A, C to D and H to G will 
complete the pattern of one-half the front piece. 



402 



Tlte New Metal Worker Pattern Book. 



The method of triangulating the piece forming 
the back of the coal hod and the development of the 
pattern of the same are so clearly shown in Figs. 
706, 707 and 70S, in addition to the plan and elevation, 
Fig. 700, as to need only a brief description. Divide 
H 1 F 1 and D" E 2 of the plan, Fig. 700, into the same 




Fig. 706. — Diagram of Sections on Solid Lines in D E F H of Fig. 100. 



number of equal parts, and from the points thus ob- 
tained erect lines vertically cutting the corresponding 
lines H F and D E of the elevation, as shown by the 
dotted lines. Connect points of like number in that 
view by solid lines and points in D E with those of 
next lower number in H F by dotted lines. Since D E, 
being inclined, is longer than M a E 2 , its equivalent 
in the plan, it will be necessary to develop an ex- 
tended section upon the line D E of the elevation, as 
shown by D 5 E 3 of the plan, which may be clone in 
the same manner as the section on the line C D above 
explained. Upon M a E 2 extended, as E 2 E 3 , set off the 



i . 



2 4 3 2 1 

5' 

Fig. 707.— Diagram of Sections on Dotted Lines in D E F H of Fig. 700. 



spaces in D E, and through the points thus obtained 
draw lines at right angles, as shown, which intersect 
with lines drawn parallel with M 2 E 3 from points of 
corresponding number in D 2 E 2 , thus establishing the 
curve D 5 E 3 , from which a correct stretchout of the top 
of the back piece may be obtained. 



In Figs. 706 and 707, the hights of the various 
points upon the perpendiculars from X and Z are equal 
to the lengths of the straight lines drawn from points 
of corresponding number in H' F 1 of the plan, Fig. 
700, to the line M 2 F'. The distances set off to the 
right upon the horizontal lines from X and Z are equal 
to the lengths of the several solid and clotted lines in 
D E F H of the elevation, and the hights of the per- 
pendiculars at the right ends of the bases are equal to 
the straight lines drawn from points in the section 
D 5 E 3 to the line E 2 E\ The several oblique solid and 
dotted lines are, therefore, the true distances repre- 
sented by the solid and dotted lines of corresponding 
number in the elevation. 




Fig. 70S. — Half Pattern of Back Piece of Funnel Coal Hod. 



In Fig. 70S, E F is equal to E F of Fig. 700 and 
is made the base of the first triangle, from which base 
the several triangles constituting the complete pattern 
may be developed in numerical order and in the usual 
manner from the dimensions obtained in Figs. 706 and 
707 and in the plan and section in Fig. 700, all as 
clearly indicated. 

The pattern for the piece forming the foot of the 
coal hod is a simple frustum of an elliptical cone, the 
method of obtaining which is fully explained in Prob- 
lem 171. In Fig. 546 of that problem the lines E F and 
G H are drawn much further apart than the proportions 
of the foot in the present case would justify, but the 
operation of obtaining its pattern is exactly the same. 



Pattern Problems. 

PROBLEM 212. 



408 



Patterns for a Three-Piece Elbow to Join a Round Pipe with an Elliptical Pipe. 



In Fig. 709, let ABCD represent the profile of 
the round pipe and E F G H I J K L the elevation of 
the elbow. In the plan the profile of round pipe is 
represented by A 1 B 1 C D 1 , the elbow by P 1 M' G' W 



PROFILE 
C 



ELEVATION 




Fig. 709.— Plan and Elevation of Three-Piece Elbow, Bound at One End and 
Elliptical at the Other. 



P 2 , and the shape of elliptical end of elbow by J 1 M 1 
G 1 M\ The section J G H I of elevation is without 
flare, and sections K F G J and EFKL are flared, as 
shown in plan. Through the plan draw E 1 G 1 and M 1 
M 2 , and carry M 1 M 2 through the center of elevation, 
as shown by M N P. Perpendiculars dropped from 
the points K F of elevation, cutting P' M 1 , E 1 G' and 
P 2 M 2 , as shown by K 1 O 1 F 1 O 2 , will give the shape of 
miter line K F in plan. 



As J G H I of elevation is without flare the pat- 
tern for this part is procured in the ordinary manner 
as for a pieced elbow. Since J 1 M 1 G 1 M 2 is the profile 
of an elliptical cylinder, the section upon the oblique 

line J G of the elevation, cut- 
ting the same, must necessarily 
be an ellipse whose major axis 
is equal to M' M 2 and whose 
minor axis is equal to J G. In 
like manner, the section at K F 
of the elevation may be assumed 
as an ellipse whose minor axis 
is K F and whose major axis is 
equal to 0' 0\ 

In Fig. 710, a duplicate of 
K F G J of elevation is shown 
by T E TJ W. Bisect T E in 
d and erect the perpendicular d 
S, and make d S equal to Q 1 O 1 
of plan. Through the points T, 
S and E trace the half ellipse, 
as shown. In a similar manner 
bisect W TJ in I, and erect the 
perpendicular I V, in length 
equal to N 1 M 1 of plan. Through 
the points thus obtained trace 
the half ellipse W V TJ. Divide 
T S E into any convenient num- 
ber of equal parts, and from the 
points thus obtained drop per- 
pendiculars cutting T E, as 
shown. Also divide "W V TJ 
into the same number of equal 
parts as was T S E, and from ' 
the points thus obtained drop 
perpendiculars cutting W TJ. 
Connect points in T E with 
those opposite in W TJ, as shown by the solid lines. 
Thus connect a with h, b with/, c with k, etc. Also 
connect the points in T E with those in W 0, as indi- 
cated by the dotted lines. Thus connect 1 with h, a 
withy, b with k, etc. 

The next step will be to construct a series of sec- 
tions upon the several solid and dotted lines just drawn 
for the purpose of obtaining the true distances which 
they represent. The sections represented by solid lines 



401 



Tlie New Metal Worker Pattern Booh. 



are shown in Fig. 711. To construct these sections 
proceed as follows : For section a, draw the line a h, 




Fig. 710.— Elevation of Middle Section of Elbow, Showing Method 
of Triangulation. 




d I 

Fig. 711.— Sections on Solid Lines of Fig. 710. 

in length equal to a h of Fig. 710. From a erect a 
perpendicular, in length equal to a 2 in T S B, as shown 



by a 2 1 , and from h erect a perpendicular, in length equal 
to h 11 in W V IT, as shown by /; ll 1 , and connect 2 1 
ll 1 , as shown. For section 5, draw b /, in length equal 
to bj of Fig. 710. From b erect the perpendicular b 3 1 , 
in length equal to b 3 in T B S, and from / erect a per- 
pendicular, in length equal to j 12 in W V U, and con- 
nect 3 1 12', etc. For the sections representing the 
dotted lines, as shown in Fig. 712, proceed in a similar 
manner. For section h, draw 1 h, in length equal to 
1 h of Fig. 710. From h erect a perpendicular, in length 



li- 



9\ 

g is 



12- 



■> / 



13- 



8*„-' 



6 2 



"~n la- 



s' 1 



15 2 



C I d m 

Fig. 712.— Sections on Dotted Lines of Fig. 710. 

equal to A 11 in W V U, and connect 1 with 11 s . For 
section/, draw a j, in length equal to a j of Fig. 710. 
From a erect the perpendicular a 2 s , in length equal to 
a 2 in T S E, and from/ erect a perpendicular, equal in 
length to/ 12 in W V U, and connect 2 2 12 2 , etc. 

To describe the pattern for part of article repre- 
sented in Fig. 710 byTEU W, as shown in Fig. 713,. 
proceed as follows : Draw any line, as E V of pattern, 
in length equal to E IT of Fig. 710. With B of pat- 



Pattern Problems. 



405 



tern as center, and 1 ll 5 of the diagram of sections, Fist. 
712, as radius, describe a small arc, which intersect with 
one struck from point V of pattern as center, and U 11 of 
Fig. 710 as radius, thus establishing the point 11 of 
pattern. With point 11 of pattern as center, and 2' 11' 
of diagram of sections, Fig. 711, as radius, describe a 



tances between points represented by numbers in Fig. 
711 for the length of solid lines in pattern, and the 
spaces in E S T for the stretchout of B T' of pattern. 
Lines drawn through the points thus obtained, as indi- 
cated by E T 1 W V, will be one-half of the required 
pattern. The other half, as shown by E T "W V, can 




Fiij. 713. — Pattern of Middle Section of Elbow. 



small arc, which intersect with one struck from E of 
pattern as center, and E 2 of Fig. 710 as radius, ttras 
establishing point 2 of pattern. Proceed in this man- 
ner, using the dotted lines in Fig. 712 for the distances 
in pattern represented by dotted lines ; the spaces in 
W V U for the stretchout of V W of pattern ; the dis- 



be obtained by a repetition of the same pi-ocess or by 
duplication. 

The pattern for E V K L of elevation can be ob- 
tained in a similar manner, A B C of profile being one- 
half the shape on E L and E S T of Fig. 710 being 
the half section on F K. 



PROBLEM 213. 



Patterns for a Right Angle Piece Elbow to Connect a Round with a Rectangular Pipe. 



In Fig. 714 is shown the design of a right angle 
elbow of which one end is rectangular, as shown by 
NOPQ, and the other rcrand, as shown by A B C D. 
Such an elbow may be constructed in any number of 
pieces, the elevation for which may be drawn in the 
manner described in the case of an ordinary piece 
elbow. 

In the present instance the elbow consists of seven 
pieces. In adjusting the transition from the rectangle 
to the circle, it is evident that the flat sides of each of 
the five intermediate pieces must become shorter in 
each piece as the round end of the elbow is approached ; 
and that the quarter circles forming the corners of each 
of the intermediate pieces must be of shorter radius as 
the corners of the rectangular end are approached. 



This may be accomplished upon the elevation in the 
following manner: Through the center of the eleva- 
tion draw the lines c to m, and divide L V into the ' 
number of parts there are pieces in the elbow subjected 
to the change in shape, in the present instance five. 
From h' set off each way four spaces, as shown b} r k Je 
and h' k". Set off from f three spaces, as shown by 
j f and/'/". Continue this operation and connect the 
points L h j h g f and g" h" j" k" L', thus showing in 
side elevation the change from the rectangle NOPQ 
to the circle A B C D. 

To show a similar shape on the outer carve of 
elbow, draw any line, as E M, in Fig. 715. From E 
on E M set off the spaces E F, F G, etc., to L M of 
Fig. 714. As it is only necessary to show the half 



406 



Tlie New Metal Worker Pattern Book. 



shapes, from M and L erect perpendiculars, in length 
equal to N n of profile, and connect same, as shown 
by M" L\ Connect 1/ F, and from the points in E M 



ELEVATION 




Fig. 714. — Elevation and Profiles of an Elbow to Connect a Round 
With a Rectangular Pipe. 



erect perpendiculars cutting F L\ The shapes on in- 
ner curve of elbow, as shown in Fig. 716, are obtained 
in the same manner as described for Fig. 715. 

For the hights of section on c m of the elevation, 
on any line, as E M, in Fig. 717, starting from E, set 
off the distances cf,fg', g' h', etc., and from the points 
thus obtained erect perpendiculars, as shown. From 
E and F set oft the distance C B of profile of circular 
end and draw E" F" From M and L set off the dis- 
tance N n of profile of rectangular end and draw M 3 
L'. Connect L 3 with F 3 , thus completing the section. 

The method for obtaining the patterns for sections 
E F F' E' and L M M' L' is the same as for an ordinary 
pieced elbow. The method for obtaining the pattern 
ior one of the remaining sections will be shown, which 



will indicate the method to be followed in the other 

sections. 

In Fig. 718, F G G-' F' is a duplicate of the sec- 
tion having similar letters in elevation. 
The shape F U F' is the half of an 
ellipse, because F' F of Fig. 714 is an 
oblique section of a cylinder of which 
A B C D is the plan, and can be de- 
scribed in any convenient manner. 

From G erect the perpendicular G 
S, equal to G G 2 of Fig. 715, and from 
G' erect another perpendicular, equal to 
G G" of Fig. 716. From points g and g' 
erect perpendiculars equal to G 3 G of 
Fig. 717, and connect T T', as shown. 
Connect S T and T' S' by a quarter of 
an ellipse. Divide S T, T' S', F U and 

U F' into any convenient number of equal parts, and 

from the points thus obtained drop perpendiculars cut- 



M°| 


M 
L 

K 


lA 


K c \ 


\ 


J 

H 


J \ 


H°\ 


gV" 


G 


^ 


F 




E 



Fig. 715.— Shape of Flat Part of 
Outer Curve of Elbow. 



Fig. 'He,— Shape of Flat Part of 
Inner Curve of Elbow. 



ting G G' and F F'. Connect these points, as shown 
by the solid and dotted lines in G G' F' F. 

The next operation will consist in constructing 
sections upon these solid and dotted lines for the pur- 



Pattern Problems. 



40 ', 



pose of ascertaining the correct distances which they 
represent. These sections are shown in Figs. 719 and 
720. To construct the sections represented by solid 




Fig. 718.- 



■Elevation of One of the Pieces, Shouting Method of 
Triangulation 



lines in Fig. 718, proceed as follows: Draw the line 
F G of Fig. 719, in length equal to F G of Fig. 718, 
and from point G erect a perpendicular, in length equal 
to G S of Fig. 718. Connect F S, which gives the 
distance between points F and S. For the second 
section draw v o, in length equal to v o of Fig. 718, 



M d 




L 3 






uA. 










\ 










r B l 




E'l 



F 

E 
Fig. 717.— Extended Section on Line c m of Fig. 714. 



and make the perpendiculars o 6 and v 7 of the section 
equal to lines having similar letters in Fig. 718. The 
other sections are obtained in a similar manner. To 



construct the sections represented by dotted lines in 
Fig. 718, proceed as follows : Draw the line F o of 
Fig. 720, in length equal to F o of Fig. 718, and from 
o erect the perpendicular o 6, equal to o 6 of Fig. 718, 
and connect F 6. For the second section draw v p, in 
length equal to vp of Fig. 718, and from points v and 
p erect perpendiculars equal to v 7 and p 5 of Fig. 718. 
Connect points 7 5, which give the distance between 
corresponding points in Fig. 718. 

In Fig. 721, GG'F'F is the pattern of part of 



/ 



1 G 1 



Fig. 719.— Sections on Solid Lines of Fig. 718. 

article shown by similar letters in Figs. 714 or 718, 
The distances represented by solid lines in pattern are 
obtained from the sections in Fig. 719, as indicated by 
corresponding figures, and the distances represented by 
dotted lines in pattern are obtained from the sections 
in Fig. 720. The stretchout of G G' of the pattern is 
obtained from GTT' G' of Fig, 718, as the stretchout 
of F U F' of the pattern is obtained from F U F' of 
Fig. 718. 



408 



Tlie New Metal Worker Pattern Booh. 



To develop the pattern from the sections above 
constructed, first draw G F of Fig. 721, in length equal 
to G F of Fig. 718 or 719, upon which duplicate the 



10 Si- 



lo 



-\0 



-i9 



11' 'S 1' 'F 

Fig. 720. — Sections on Dotted Lines of Fig. 718. 

triangle G F S of Fig. 719, as shown. From S of pat- 
tern as center, with S 6 of Fig. 718 as radius, de- 
scribe an arc, 6, which intersect with one struck from 



F of pattern as center, with radius F 6 of Fig. 720, 
thus establishing point 6 of pattern. Then with radius 
F 7 of Fig. 718, from F of pattern as center, describe 
an arc, which intersect with a second arc struck from 
point 6 of pattern as center, and 6 7 of Fig. 719 as 
radius, thus establishing point 7 of pattern. Continue 
this process until the various points indicated in pat- 
tern are located. Lines drawn through the points 



1 S G' 




Fig. 721— Half Pattern of F G G' F' of Fig. 714. 

thus obtained, as indicated by GSTT'S'G' and F' 
U F, will complete one-half of the required pattern. 
The other half can be obtained by duplication or by a 
repetition of the above process. 

In obtaining the pattern for any one of the re- 
maining pieces first draw a duplicate of its elevation as 
taken from Fig. 714, upon either side of which con- 
struct the proper section, obtaining the points in the 
same from Figs. 715, 716 and 717 as was done in Fig. 
718, the subsequent operations being explained in Prob- 
lem. 



PROBLEM 214. 

Pattern for the Soffit of a Semicircular Arch in a Circular Wall, the Soffit Being: Level at the Top and 
the Jambs of the Opening; Being: at Rig:ht Angles to the Walls in Plan. Two Cases. 



First Case. — In Fig. 722, let A B C represent the 
outer curve of an arch in a circular wall corresponding 
to A' H C of plan, and let E B D represent the inner 
opening in the wall, as shown by E' F' D' in plan. 
Then A E B C D will represent the soffit of the arch 
m elevation and A' H C D' F' E' the same in plan. 
In the engraving the outer curve of the arch is a per- 
fect semicircle, and the inner curve is stilted, as 
shown, so as to make the soffit level at B. Instead 
of the stilted arch, the inner curve may, if desired, be 
drawn as a semi-ellipse of which E D is the minor 
axis and F B one-half of the major axis. 

Divide A B of elevation into any convenient num- 
ber of equal parts, shown by the small figures. With 



the T-square parallel with the center line B B', drop 
lines from the points in A B, cutting A' H of plan, 
as shown. Since that portion of the inner arch from 
E to 13 is drawn vertical, as above explained, divide 
12 B into the same number of parts as was A B, and, 
with the T-square parallel with the center line B B', 
drop lines to E' F', as shown. Connect opposite 
points in A' H with those in E' F', as shown by the 
solid lines in plan. Also divide the four-sided figures 
thus produced by means of the diagonal dotted lines 
6 8, 5 9, etc., as shown. The several triangles thus 
produced will represent in plan the triangles into which 
the soffit, or under side, of the arch is divided for the 
purpose of obtaining its pattern. In order to ascer- 



Pattern Problems. 



409 



tain the real distances across the surface of the arch 
which the solid and clotted lines represent, it will be 
necessary to construct a series of sections of which 
these lines are the bases, as shown in Figs. 723 and 
724. 

In constructing the diagram shown Fig. 723, the 
several solid lines of the plan, though not exactly 
equal in length (because they are not drawn radially 
from the center of the curve A' II C), may be con- 
sidered as of the same length. Draw the right angle 
PQE as in Fig. 723, and from Q set off horizontally 
the distance H F' of plan, as shown by Q R. Draw 
R S parallel with Q P, and, measuring from Q, set off 
on QP the length of lines dropped from points in A 

ELEVATION 

B 




Fig. 722.— Plan and Elevation of Arch in a Circular Wall- 
First Case. 

B to A F, as shown by corresponding figures 2 to 6. 
Likewise set off from R on R S the length of lines 
dropped from points in E B to E F, as shown by the 
figures 12 to 7, and connect the points in P Q with 
those in S R, as indicated by the solid lines in plan. 
Thus connect 1 with 12, 2 with 11, 3 with 10, etc. 
To construct the diagram based upon the dotted lines 
of the plan, draw the right angle MNO in Fig. 724, 
and, measuring in each instance from N, set off on N 
M the same distances as in Q P of Fig. 723. Starting 
from N, set off on N" the lengths of dotted lines in 
plan, as shown by the small figures in N 0. With 
the T-square parallel with M N, draw lines from the 



points in 1ST 0, and, in each instance measuring from 
N" 0, make these lines of the same length as lines of 
similar number dropped from points in E B of eleva- 
tion to E F. Connect the points in these lines with 
points in M N, as indicated in plan by the dotted 





Fig. 723.— Diagram of Sections on Fig. 724.— Diagram of Sections on 
Solid Lines of Plan, Fig. 722. Dotted Lines of Plan, Fig. 722. 



lines. Thus connect 6 with 8, 5 with 9, 4 with 
10, etc. 

The next step is to obtain the distances between 
points in A B of elevation as if measured on the outer 
opening in the curved wall. To do this, on F A ex- 
tended set off a stretchout of A' H of plan, as shown 
by the small figures 5', 4', etc., and with the T-square 
at right angles to the stretchout line J F, draw the 
usual measuring lines. With the T-square parallel 
with J F, carry lines from the points in A B to lines 
of similar number drawn from the stretchout line. A 
line traced through these points, as shown by J B, will 
give the true distances desired between the points in 
the outer curve of the arch. 

The distances between points in E B, the inner 
curve, are obtained in a similar manner. To avoid a 
confusion of lines, the stretchout of F' E' of plan is 




13 12 1' 

Fig. 725.— One-Half Pattern of Soffit of Arch Shown in Fig. 722. 

set off on F C of elevation, as shown by the small 
figures, 7', 8', 9', etc. B D is also divided into the 
same parts as was E B, and from the points thus ob- 
tained lines are drawn to the right parallel with F C. 
With the T-square parallel with B F, carry lines 



410 



The New Metal Worker Pattern Book. 



from the stretchout points in F K, cutting lines of 
similar number drawn from the points in BD. A 
line traced through the points thus obtained, as shown 
by B K, will give the distance between points as if 
measured on the inner curved line of the wall. 

From the several sections now obtained the pat- 
tern may be developed in the following manner : At 
any convenient place draw the line a e in Fig. 725, 
making it in length equal to Q R of Fig. 723, or A' E' 



thus establishing the point 11 of pattern. Continue 
in this way, using the tops of the sections in Figs. 
723 and 721 for measurements across the pattern, the 
spaces in J B for the distances along the edge a h of 
pattern, and the spaces in B K for the distances along 
the inner edge e /, establishing the several points, as 
shown. Through the points in a h and e f lines are 
to be traced, while fh is to be connected by a straight 
line, thus completing one-half the pattern. The other 




Fig. 726.— Plan and Elevation of Arch in a Circular Wall. — Second Case. 



of plan. At right angles to a e draw e 12, in length 
equal to R 12 of Fig. 723, and connect a with 12 if it 
is desired to show the triangle. From a as center, and 
J 2 of elevation as radius, describe a small arc, 2, 
-which intersect with one struck from point 12 of pat- 
tern as center, and 12 2 of Fig. 724 as radius, thus es- 
tablishing the point 2 of pattern. With 2 of pattern 
as center, and 2 11 of Fig. 723 as radius, describe 
another arc, 11, which intersect with one struck from 
12 of pattern as center, and 12 11 of B K as radius, 



half of pattern can be obtained by the same method 
or by any convenient means of duplication. 

If the arch were semi-elliptical instead of semi- 
circular, the method of procedure would be the same 
as above described. 

Second Case. — In Fig. 726, ABC represents the 
outer curve of an arch in circular wall, as shown by 
A' H C in plan. EBD represents the inner curve 
in elevation, as does E' D' the same in plan. Then 
A E B D C represents the soffit of the arch in eleva- 



Pattern Problems. 



411 



tion and A' H C D' E' the same in plan. The con- 
ditions given in this case differ from those of the first 
case only in the fact that the inner curve of the arch 
in this case is straight in plan, as shown by E' D', 
instead of curved to the radius of the wall as in Fig. 
722. The method of procedure in this case is exactly 
the same as before, but one less operation will be nec- 
essary, since measurements upon the inner curve may 
be taken directly from E B D of the elevation. 

To avoid a confusion of lines, a duplicate 





Fig. 727. — Diagram of Sections on 
Solid Lines of Plan, Fig. 726. 



Fig. 72S. — Diagram of Sections on 
Dotted Lines of Plan, Fig. 726. 



of EBD of elevation has been drawn in plan, as 
shown by E' B' D'. To obtain the divisions on plan 
divide A B into any convenient number of equal parts, 
and from the points thus obtained drop lines parallel 
with the center line B B' to A' H of plan, as shown. 
Divide E' B' in a similar manner, and from the points 
thus obtained drop lines to E' F' of the plan, as shown. 
Connect points in A' H' with those of similar number 
in E' F' by solid lines. Also connect points in A' H 
with those of next lower number in E' F' by clotted 
lines. These solid and dotted lines just drawn will 
form the bases of a series of sections, shown in Figs. 
727 and 728, whose upper lines will give correct dis- 
tances across the pattern of the soffit. 

To construct the sections based upon the solid 
lines of the plan, first draw the right angle P Q B in 
Fig. 727, and set off on Q P, measuring from Q, the 
length of the vertical lines in A B F of elevation. 
Starting from Q, set off on Q R the length of solid 
lines in A' H F' E' of plan, as shown by the small 
figures in Q R. With the T-square parallel with P Q, 
draw lines from the points in Q R, and, measuring 
from Q R, set off on these lines the length of lines of 
corresponding number in E' F' B' of plan, and connect 
the points with points of similar number in P Q. The 
diagram of sections based upon the dotted lines of the 
plan, shown in Fig. 728, is constructed in the same 



manner, using the length of dotted lines in plan for 
the distances in N 0, the length of lines in E' F' B' of 
plan for the length of lines set off at right angles to 
N 0, and the length of lines in A B F of elevation 
for the distances in M N. Connect the points as in- 
dicated by the dotted lines of the plan, all as shown. 

The next step is to obtain the correct distances 
between points in A B of elevation, or A' H of plan. 
To do this lay off horizontally J K, on which set off 
a stretchout of A' H of plan, and, with the T-square 
at right angles with J K, draw the usual measuring 
lines. With the T-square parallel with J K, carry 
lines from the points in A B to lines of similar num- 
ber. A line can be traced through these points, as 
shown by J L, from which the correct stretchout of 
the outer side of the pattern can be obtained. 

To describe the pattern first draw any line, as a e 
of Fig. 729, equal to A' E' of plan. With e of pattern 
as center, and E' 1 of the inner curve of the arch as 
radius, strike a small arc, 1', which intersect with one 
struck from a of pattern as center, and Q 1' of Fig. 
727 as radius, thus establishing the point 1' of pat- 
tern. With a of pattern as center, and J 2 of Fig. 

726 as radius, describe a small arc, 2, which intersect 
with one struck from 1' of pattern as center, and V 2' 
of Fig. 728 as radius, thus establishing point 2 of 
pattern. Then from 2 as center, with 2' 2' of Fig. 

727 as radius, strike a small arc, 2', which is inter- 




s' -y 

Fig. 729.— Half Pattern of Soffit Shown in Fig. 726. 

sected with one struck from 1' of pattern as center, 
and 1 2 of the inner curve of the arch as radius, thus 
definitely establishing the point 2' of pattern. Con- 
tinue in this way, using the tops of sections in Figs. 
727 and 72S for measurements across the pattern, and 
the spaces in the inner curve E' B' and in outer curve 
as developed in J L, Fig. 726, for the distances about 
the edges of the pattern, establishing the several 
points, as shown. Connect points h and f with ", 
straight line, and through the intermediate points trace 
the curves, as shown. The other half of pattern can 
be obtained by the same method or by duplication. 



4:12 Tlie New Metal Worker Pattern Book. 

PROBLEM 215. 

Pattern for a Splayed Elliptical Arch in a Circular Wall, the Opening: Being- Larg-er on the Outside 

of the Wall than on the Inside. 



In Fig. 730 is shown the elevation and plan of 
an elliptical window head in a circular wall. The 
outer curve of head is represented by A B C in elevation 
and by A' B' C in plan. The inner curve is repre- 
sented by F E D in elevation and F' E' D' in plan. 
A B C D E F therefore represents the splayed or 
beveled portion in elevation for which the pattern is 
required, and A' B' C D' E' F' the plan of the same. 
Divide A B of elevation into any convenient number 



of sections on A' B' of plan, as the lines dropped from 
F E to F G give the hight of sections on F' E' oi 
plan. 

To construct the sections shown in Fig. 731, rep- 
resented in plan by the solid lines, proceed as follows : 
Draw the right angle a a' 1, making a a' equal to E' B' 
of plan, and a' 1 equal to G B (a 1) of elevation. Draw 
a 12 parallel to a 1, making its length equal to G E 
(a 12) of elevation, and connect 12 with 1. The dis- 





L g e a b 

DEVELOPMENT OF 
INNER CURVE, 



Fig. 7S0. — Elevation and Plan of Splayed Elliptical Arch. 



of parts, in the present instance five. For convenience 
number the points thus obtained, as shown by the 
small figures. Drop perpendiculars from the points 
in A B, cutting A' B' of plan, as shown. Also divide 
F E of elevation into the same number of parts as was 
A B, and drop similar perpendiculars to F' E' of plan. 
Connect opposite points in A' B' with those in F' E', 
as shown by the solid lines, as 2 11, 3 10, etc. Also 
divide the four-sided figures thus produced into tri- 
angles by means of the dotted lines, as shown from 1 
to 11, 2 to 10, etc. To ascertain the true distances 
across the face of the arch which these several solid 
and dotted lines of the plan represent, it will be nec- 
essary to construct vertical sections through the arch 
upon each one of these lines as a base. The lines 
dropped from A B to A G of elevation give the hight 



tance 12 1 of this section represents the actual distance 
between the points 1 and 12 in the elevation or plan. 
The second section is constructed in a similar manner; 
h c represents the distance 2 1 1 of plan ; b 11 is equal 
to b 11 of elevation and c 2 to c 2 of elevation. Con- 
nect 11 with 2 of section, which will give the actual 
distance between points 11 and 2 of elevation or plan. 
The remaining sections are constructed in a similar 
manner, each of the sections representing a vertical 
section through the head on the lines of corresponding 
numbers in the plan. The sections based upon the 
dotted lines of the plan, shown in Fig. 732, are con-, 
structed in exactly the same manner. Draw b a, in 
length equal to 1 11 of plan, and erect the two perpen- 
diculars, as shown. Make a 1 of section equal to a 1 
of elevation, and also make b 11 of section equal to b 



fattem froblems. 



413 



11 of elevation, and connect 11 with 1. The remain- 
ing sections are constructed in the same manner. 

Before the pattern can be obtained it will be nec- 
essary to develop extended sections of the inner and 



«/< 




a b 




e j a Ji 

Fig. 731. — Sections Based Upon the Solid Lines of the Plan, Fig. 780. 

outer curves, as shown to the left and right of the ele- 
vation. This is clone for the purpose of obtaining the 
actual distance between points shown in elevation. 
For convenience, on G A extended, as H K, lay off a 
stretchout of A' B' of plan, and from the points therein 
contained erect the perpendiculars, as shown. From 
the points in A B of elevation draw lines parallel with 
H G, cutting perpendiculars of similar number erected 
from H K. A line drawn through these points of in- 



to / 



7^: 



Fig. 732. — Sections Based Upon the Dotted Lines of the Plan, Fig. 730. 

tersection, as shown by H J, will show the shape of 
A B of elevation as laid out on a flat surface. The 
development of the inner curve is shown to the right 



of the elevation. On LN is laid out the stretchout of 
F' E' of plan, and on the perpendiculars erected from 
the points in the line are set off the same distances as on 
lines of similar number in F E Gr of elevation. A line 
traced through these points, as shown by L M, also 
shows the shape of F E, as laid out on a flat surface, 
and gives the distance between points as if measured 
on the finished article. 

To obtain the pattern, using the distances between 
points in H J and L M of Fig. 730, and the diagrams 
in Figs. 731 and 732, proceed as follows : Draw any 
line, as B E in Fig. 733, in length equal to 1 12 of the 
first section in Fig. 731. With E as center, and M 11 
of inner curve as radius, describe a small arc, 11, which 
intersect with one struck from B as center, and 1 11 of 
Fig. 732 as radius, thus establishing the point 11 of 
pattern. "With 11 of pattern as center, and 11 2 of 
the second section in Fig. 731 as radius, describe an- 




Fig. 733. — Pattern of Splayed Arch in Fig. 730. 

other small arc, 2, which intersect with one struck 
from point B of pattern as center, and J 2 in J H of 
outer curve as radius, thus establishing the point 2 of 
pattern. Continue in this way, using the tops of the 
sections in Figs. 731 and 732 for the measurements 
across the pattern, the spaces in the inner curve L M 
and in the outer curve H J for the distances about the 
edges of the pattern, establishing the several points, 
through which draw the lines shown. Then B A F E 
is the half pattern of splayed head, shown in elevation 
by A B E F. The other half can be obtained by any 
convenient means of duplication. 

A semicircular splayed arch can be developed in 
the same manner as above described. 

The pattern for a blank for a curved molding, 
either semicircular or semi-elliptical, for an arch in a 
circular wall comprises really the same relations of 
parts as are shown in Fig. 730, and could be obtained 
as above described. 



414 



TJie New Metal Worker Pattern Book. 



PROBLEM 216. 
Pattern for a Splayed Arch in a Circular Wall, the Larger Opening: Being" on the Inside of the Wall. 



In Fig. 734 is shown the plan and elevation of an 
arch in a curved wall, such as might be used as the 
head of a window or door, the jambs and head to have 
the same splay. In the plan, CED represents the 
inner curve of wall and A F B the outer. J G M in 
elevation represents the inner curve and K H L the 
outer. In order to arrive at a system of triangles by 
means of which to measure the splayed surface, first 



the inner and outer lines of plan, as shown. The solid 
and dotted lines of the plan will form the bases of a 
series of right angled triangles whose altitudes can 
be found in the elevation, and whose hypothenuses 
will give correct measurements across the face of the 
splayed surface. 

To determine the hights of these triangles extend to 
the left the horizontal lines drawn throusrh K H of the 



ELEVATION 
G 




DEVELOPMENT OF 
OUTER CURVE 



Fig. 



PLAN 

7S4. — Plan, Elevation and Extended Sections of Splayed Arch. 



divide J G of the elevation into any convenient num- 
ber of parts, in this case five, as indicated by the small 
figures. From the points in the curve thus established 
drop lines parallel to the center line G F, cutting the 
inner curve of the plan C E, as shown. Next carry 
lines from points in J G in the direction of the center 
X, intersecting the outer curve and establishing the 
points 7, 8, 9, etc., in it. From these points drop lines 
to the outer curve in plan A F, establishing the points 
7, 8, 9, etc. Connect opposite points in J G with 
those in K H, as 1 and 7, 2 and S, 3 and 9, 4 and 10, 
5 and 11, and 6 and 12. Likewise connect 1 and 8, 
2 and 9, 3 and 10, etc., as shown by the dotted lines. 
In the same manner connect corresponding points in 



elevation until they intersect lines dropped from J G, 
as shown by the points b, d, ,/and h. Then b 2, d 3, 
etc. , will be the required hights. To construct the tri- 
angles which represent the solid lines in plan and ele- 
vation, proceed as shown in Fig. 735 : For the first 
triangle, G H 7, draw the right angle G H 7, making 
the altitude equal to G H of elevation and the base 
equal to E F, 1 7, of plan, and draw the hypothenuse, 
which represents the actual distance between the points 
1 and 7 of the elevation or plan. In like manner the 
hypothenuse of the second triangle 2 5 8 shows the 
actual distance between the points 2 and 8 of the 
elevation; 2 b in Fig. 735 is made equal to 2 b of the 
elevation, while b 8 equals 2 8 of the plan. 



Pattern Problems. 



415 



The remaining triangles in Fig. 735 are con- 
structed in a similar manner, each of the triangles rep- 
resenting a vertical section through the head on the 
lines of corresponding numbers in the plan. 

The triangles shown in Fig. 730 correspond to 
similar sections taken on the dotted lines in plan. The 
base a S of the first triangle is equal to 1 S of the plan, 
and the hight la to 1 a of elevation, the point a of 
the elevation being on a level with 8, as shown by 
the dotted line 8 a. The hypothenuse 1 S is then 
drawn, which gives the distance between the points 1 




s H 





Fig. 735.— Diagram of Triangles Based Upon the Solid Lines of 
the Plan. 

and 8 in plan or elevation. The bases of the remain- 
ing triangles are derived from the dotted lines in plan, 
and the hights from the distances 2 c, 3 e, ± g and 5 k 
of elevation. 

Before the correct measurements or stretchouts 
for the inner and outer lines of the pat- 
tern can be obtained it will be neces- 
sary to develop extended sections of the 
inner and outer curves of the arch in 
elevation, as shown at the left and 
right of that view. For the develop- 
ment of the outer curve, as shown to 
the right by A' II' F', proceed as fol- 
lows : On X M extended lay off a 
stretchout equal to A F of the plan, 
transferring it point by point. From 
the points thus established in A' F' 
carry lines vertically, extending them 
indefinitely, as shown, and then from the points in the 
outer curve K II of the elevation carry lines horizon- 
tally to the right, intersecting the corresponding lines 
just drawn from A' F', and through the points thus 
established trace the curved line A' H'. The de- 



velopment of inner curve, as shown to the left, is 
accomplished in a similar manner. On X J ex- 
tended lay off a stretchout of C E of plan, and from 
the points thus established carry lines vertically. 
From the points in the inner curve J G carry lines 
horizontally to the left, intersecting lines of similar 
number, and through the points thus established trace 
the curve C G'. 

To describe the pattern shown in Fig. 737, first 



\ 



\ 



_\. 





Fig. 736.— Diagram of Triangles Based Upon the Dotted Lines of 
the Plan. 

draw the line g h, or 1 7, in length equal to the hypoth- 
enuse 1 7 of the first triangle in Fig. 735. From 7 as 
center, with 7 8 of the development of the outer curve as 
radius, strike a small arc, as shown at 8 in the pattern. 
From 1 as center, with 1 8 of the first triangle in Fig. 




PATTERN 
Fig. 737.— Pattern for Splayed Arch. 



736 as radius, intersect the arc last struck, thus estab- 
lishing the point 8. From 1 as center, with radius 
equal to 1 2 of the development of inner curve, strike 
a small arc, as shown at 2. Then from 8 as center, 
with 8 2 of the second triangle in Fig. 735 as radius, 



416 



The New Metal Worker Pattern Book. 



intersect the arc at 2 already drawn, thus definitely 
establishing the point 2 in the upper line of the pattern. 
Continue in this way, using the hypothenuses of the 
several triangles, as shown, for measurements across the 
pattern, and the spaces of the inner and outer curves 



as developed in Fig. 734 for the distances along the 
edges of the pattern, establishing the several points, as 
shown. Lines traced through the points from c to g 
and from a to h will complete the pattern for one-half 
the arch. The complete pattern is shown in Fig. 737. 



PROBLEM 217. 

Pattern for the Soffit of an Arch in a Circular Wall, the Soffit Being: Level at the Top and the 
Jambs of the Opening Being- Splayed on the Inside. 



In Fig. 738, A B G is the elevation of the inner 
curve A' H C of the plan and £ B D that of the outer 
curve E' F' D'. As will be seen by inspection, the 
outer curve is drawn from G as center, that portion of 
the opening from the springing line down to the spring- 



the outer curve of wall is struck from G as center, 
divide 12 B into the same number of parts as was A B, 
and drop lines from these points to E' F' of plan, as 
shown. Connect opposite points in E' F' of plan with 
those in A' H, as indicated by the solid lines. Also 




M 1211 10 9 



7U 



Fig. 7SS. — Plan, Elevation and Extended Sections of an Arch in a 
Circular Wall. 



SHAPE ON E F'oF PLAN 



ing line of the inner curve, as E 12, being straight and 
vertical. The pattern of the soffit could have been 
obtained in exactly the same manner had the elevation 
of this outer curve been a semi-ellipse. 

Divide A B of the elevation line into any con- 
venient number of equal parts, and with the T-square 
parallel with center line B H drop lines from the 
points in A B, cutting A' H of plan, as shown by the 
small figures 1 to 6. As the semicircle representing 



connect the points of the plan obliquely, as shown by 
the dotted lines, thus dividing the plan of the soffit of 
the arch into triangles. In order to ascertain the true 
distances which these lines drawn across the plan repre- 
sent it will be necessary to construct a series of sections 
of which they are the bases, as shown in Figs. 739 and 
740. 

In Fig. 739 is shown a diagram of sections corre- 
sponding to the solid lines in plan, to construct which 



Pattern Problems. 



417 



proceed as follows : Draw the right angle P Q R, 
and, measuring from Q, set off on Q P the length of 
lines dropped from points in A B of elevation to A F, 
as shown by the figures 2 to (3. Likewise set off from 
Qon QR the length of solid lines in plan, as shown by 
the small figures 7 to 12. With the "[-square parallel 
with P Q, erect lines from the points in Q R, and, 
measuring from Q R, set off on these lines the length 
of lines of corresponding number in E B F of elevation, 




----. 10 I 

- — 41 

~H11 
II 

" — U_ ,12 

! 
i 



u 



Fig. 7S9. — Diagram of Sections on Fig. 740. — Diagram of Sections on 
the Solid Lines of the Plan. the Dotted Lines of the Plan. 



and connect the points thus obtained with the points 
in P Q, as indicated by the numbers in the plan. Thus 
connect 6 with 7, 5 with S, 4 with 9, etc. 

To construct a diagram of sections corresponding 
to the dotted lines of the plan draw the right angle 
S T U, Fig. 710, and set off on S T the length of lines 
dropped from points in A B of elevation to A F, or 
tranfer the distances in P Q, Fig. 739. On T U set 
off the length of dotted lines in plan, and from the 
points thus obtained draw lines parallel with S T, 
making these lines of the same length as those dropped 
from points in E B of elevation to E F. Connect the 
upper ends of lines 8 to 12 with points in S T, as indi- 
cated by the dotted lines in plan. Thus connect 2 
with 12, 3 with 11, 1 with 10, etc. 

The next step is to obtain the distance between 
points in A B and E B of elevation as if measured upon 
the curved surfaces of the wall. It is therefore neces- 
sary to develop extended sections of the two curves of 
the arch as shown at the left in the enarravins;. The 
development of the curve of the inner side of the arch 
is projected directly from the elevation in the follow- 
ing manner : On A F extended lay off J K, equal to 
the curved line A' H of plan, making the straight line 
equal to the stretchout of half of the curve of the plan, 
as indicated by the small figures. At right angles to 
J K, and from the points in the same, erect lines, 



making them the same length as lines of similar num- 
ber dropped from points in A B, or, with the "[-square 
parallel with J K and A F, carry lines from the points 
in A B intersecting the vertical lines of similar num- 
ber. A line traced through the points of intersection, 
as shown by JL, will'be the desired shape. The shajDe 
on E' F' of the plan, corresponding to E B F of eleva- 
tion, is obtained in a similar manner. On M N in Fig. 
738 set off the stretchout of E' F' of plan, as indicated 
by the small figures, from the points of which erect 
vertical lines. On these lines set off the same length 
as the lines of similar number in E B F of elevation. 
A line traced through the points thus obtained will 
give the desired outer curve of the arch. 

To develop the pattern from the several sections 
obtained proceed in the following manner : Draw the 
line a e of Fig. 741, in length equal to Q R of Fig. 
739, and with e as center, and M 12 of the curve M O 
as radius, strike a small arc, 12, which intersect with 
one struck from a as center, and Q 12 of Fig. 739 as 
radius, thus establishing the point 12 of pattern. 
With a of pattern as center, and J 2 in the curve J L 
as radius, describe a small arc, 2, which intersect with 
one struck from point 12 of pattern as center, and 2 12 
of Fig. 740 as radius, thus establishing the point 2 of 
pattern. With 2 of pattern as center, and 2 11 of 
Fig. 739 as radius, strike another small arc, which 
intersect with one struck from point 12 of pattern as 
center, and 12 11 in the curve M as radius, thus 
establishing the point 11 of pattern. Continue in this 
way, using the tops of the sections in Figs. 739 and 




2 3 4 

Fig. 741— Half Pattern of Soffit. 

740 for measurements across the pattern, and the 
spaces in the inner and outer curves as developed in 
J L and M O, Fig. 738, for the distances about the 
edges of the pattern, establishing the several points, 
as shown, through which draw the lines e f, a h and 
/ h\ thus completing the pattern for one-half the soffit 
of the arch. The other half of pattern can be ob- 
tained by the same method or by duplication. 






418 



The New Metal Worker Pattern Book. 

PROBLEM 218. 



Pattern of the Blank tor a Curved Molding in an Arch in a Circular Wall. 



In the last paragraph of Problem 215 it is stated 
that the demonstration there given is applicable to the 
blank for a curved molding in a circular wall. There 
are many forms of arches and different methods of 
adapting them to the requirements of a curved wall. 
An arch may be semicircular or elliptical, having 
either the long or the short diameter of the ellipse as 
its width ; and in either case it may be rampant, 
though rampant arches are seldom used. Any form 
of arch may be so constructed that its moldings project 
either from the exterior or from the interior surface 
of a curved wall. In adapting any form of arch to a 
circular wall the soffits and roof strip, or portions which 
appear level at the top of the arch, may, as they are 
carried around the curve of the arch, arrive at the 
springing line or top of the impost parallel to the cen- 
ter line in plan ; or they may at this point be drawn 
radially to the curve of the wall. 

In Fig. 7-42 is shown one-half of the elevation 
and plan of a semicircular window cap and a section 
on the center line of the same. M K L 1ST is one- 
half the elevation of all the members constituting the 
molding, the lines of which are projected from the 
profile W as shown to the right of the center line. In 
the plan, those lines which in profile W were drawn 
horizontally are here drawn parallel to the center line 
B G. They might with equal propriety have been 
drawn radially toward the center of the curve E G. 
Should they be drawn radially the profile of the mold- 
ing would remain nearly normal throughout its course, 
but when drawn as in the plan, Fig. 742, it will be seen 
that the profile is continually changing, that at the 
foot of the arch or top of the corbel being shown at Y 
in the plan. This profile is obtained by the usual 
operation of raking, as shown by the dotted lines. 

Aa the blank for any curved molding is, to the 
pattern cutter, a flaring strip of metal, it simply be- 
comes necessary to determine its width and the amount 
of flare which it may assume in the different parts of 
its course, after which its pattern may be arrived at 
by methods described in Problem 214 and those fol- 
lowing. The direction of the line determining the 
amount of flare necessary for the strip to have is de- 
termined by the judgment of the pattern cutter and 
the requirements of the machinery used in "raising" 



the mold. Therefore in making the application of 
the demonstration given in Problem 215 to the win- 
dow cap shown in Fig. 742 it is necessary to first 
draw the lines k g of the profile in elevation, and p r 
of the plan, establishing the flare and necessary width 
or stretchout at those points, after which the points 




J?ig. 742.— Elevation and Plan of an Arched Window Cap in a 
Circular Wall. 



p and r may be dropped upon the line M N of the 
elevation, and the points k and g carried horizontally 
to the center line K X, as shown. The points thus 
obtained in M N and K L must then be connected 
by the necessary arcs struck from X as center, thus 
completing an elevation of the flaring piece. Likewise 
the projection of the points g and k from K L must 
be set off from C on C G of the plan, as shown at g" 
and k", and the points thus obtained connected with 



Pattern Problems. 



419 



r and p by arcs struck from the center used in describ- 
ing the plan of the wall. 

Should the width and flare of the blank, as deter- 
mined by the raked profile Y at the foot of the arch, 
vary so much 'from those of the normal profile at the 
top that parallel curved lines could not be drawn to 
connect the points at the foot with those at the top in 
either or both views, then centers must be found upon 



730 ; after which the demonstration given in that 
problem may be followed to obtain the required 
pattern. 

It will thus be seen from the foregoing that no 
matter what form of arch be used or in what manner 
it be placed upon the wall, the method of obtaining 
the pattern of its flaring surfaces remains the same. 

It might under some circumstances be desirable 







Fig. 74-3.— Method of Obtaining a Section through Molding. 



the center lines of the plan and of the elevation from 
which arcs can be drawn connecting the required 
points. Having thus completed a plan and elevation of 
the flaring piece or blank, it will be seen by reference 
to Problem 215 that the lines p' r' and 7/ g' of the 
elevation, Fig. 742, correspond with A F and B E of 
Fig. 730 ; and that p r of the plan, Fig. 742, and the 
arcs drawn from it correspond with B' A' F' E' of Fig. 



to construct stays at several intervals between the top 
and the foot of an arch, similar to that shown at Y in 
Fig. 742, for the purpose of more accurately determin- 
ing the flare in all parts of its sweep, or for the purpose 
of constructing a form of templet to assist in the 
operation of raising the mold. The profiles of such 
stays can be obtained by the usual operation of rak- 
ing, which is fully described in numerous problems in 



420 



The New Metal Worker Pattern Booh. 



the first section of this chapter. In Fig. 743 the 
operation of obtaining a profile upon the line X' X" is 



graving must be obtained by developing extended 
sections of those lines on E C of the plan, as de- 




Fig. 744.— Perspective View of Templet for Use in Raising the Curved Mold. 



carried out in all its detail, resulting in the form 
shown at Q, and needs no further comment. 

In constructing a form or templet, as shown in 
Fig. 744, the outlines of the arch shown in the en- 



scribed in Problem 214 and those following, after 
which the stays can be placed upon lines drawn to 
correspond with those from which the respective sec- 
tions were taken. 



PROBLEM 219. 



Details of Patterns for Automobile Dust Pan. 



This problem explains how to lay out the patterns 
for an automobile dust pan and how the intersecting 
lines between the side pocket and pan are projected, and 
how to place them on the patterns. The problem pre- 
sents an interesting study in projections and devel- 
opments. In Fig. 745, A B represents the center line, 
5 1 10' H the one half plan of the pan, showing the full 
length and the one half width. Below the plan is shown 
the one half front view, the outer curve F 14 represent- 
ing the section on line 6 14 in plan, while G 10 11 14 
shows the section on H 10' in plan. 

Above the plan is shown the one half rear view, 
the outer curve 6 D representing a section on 6 14 in 
plan and is similar to F 14 in the half front view, and 
the inner curve 5 D in the rear view representing the 
section on the line 5 1 in plan. In the rear view D e d b 
shows the head of the side pocket which, in this case, is to 
intersect with the pan 



To the right of the half front view is the side view 
of the pan and pocket, the length being a projection 
from the plan by the quadrants struck from the center 
14 in the front view, as shown by dotted lines, while 
the vertical nights are obtained from the half front 
view and rear view Z, which is a reproduction of the 
half rear view above the half plan. Thus 10 D° 5 6 G° 
shows the side view of the pan, the rear part being longer 
than the front as shown, and X F° F° D° showing the 
side view of the pocket which is to intersect the pan 
at its sides and bottom. 

As shown by the half plan, the triangular gore in- 
dicated by 10 10' 14 is to have a profile on the line 

10 10', similar to the shape shown by 10 14 in the one 
half front view. The loreshortened view of the curve 

11 14 in front view on the line 11 14 in plan, is pro- 
jected to the side view as follows: Divide the curved 
part in the half front view in equal spaces as from 11 



Pattern Problems. 



421 



to 14, from which points erect vertical lines until they 
intersect the line 10 14 in plan. From these intersec- 
tions parallel to H 10' draw lines intersecting the side 
10' 14 at 11', 12', and 13'. Using 14 in the front view 
as center, with radii equal to 14 10', 14 11', 14 12', 
14 13', and 14 14, draw the quadrants intersecting the 
top line of the pan in the side view at 10, 11, 12, 13, 
and 14. From these intersections draw vertical lines, 



jections to points 11, 12, 13, and 14 on the line 10 14, 
and place these distances on similar numbered lines, 
measuring in each instance from the line L M. Trace 
a line through points thus obtained and 10 N 14 will 
be the pattern shape. 

To obtain the line of intersection between the rear 
head of the pocket D° F° in the side view with the rear 
part of the pan, as shown in the rear view (Z) by D E ( 




Fig. 745. — Half Plan, Half Front and Rear Views, Side View True Lengths and Pattern Shapes. 



which intersect by lines drawn from similar numbers 
in the half front view, parallel to 14 X, thus obtaining 
10°, 11°, 12°, 13°, and 14, through which the dotted 
line is traced, completing the side view. 

The pattern for the triangular gore is obtained as 
follows: Take the stretchout of the profile 10 to 14 
in the half front view and place it on the horizontal 
line L M, as shown by similar numbers, through which 
erect perpendicular lines as shown. Measuring from 
the line H 10' in the half plan, take the various pro- 



proceed as follows: Starting from the intersection be- 
tween the inner and outer curves in the half rear view 
(Z) indicated by 1, divide both curves into an equal 
number of spaces, as shown by the small figures from 
1 to 5 and 6 to 1. Connect opposite points as from 
5 to 7 to 4 to 8 to 3 to 9 to 2 and from 9 to 1. From 
the various intersections on the inner curve 1 to 5, draw 
horizontal lines (not shown) until they intersect the 
proper line D° 5 in the side view, as indicated by the 
figures 1 to 5. In a similar manner from the various 



422 



The New Metal Worker Pattern Book. 



intersections on the outer curve 6 to 9 to 1, draw hori- 
zontal lines (not shown) until they intersect the proper 
line 14 6 in the side view as also indicated by the small 
figures 6 to 9 to 1. Connect opposite numbers by dotted 
lines hi a similar manner as was done in the half rear 
view (Z), as shown from 5 to 7 to 4 to 8 to 3 to 9 to 2 
and from 9 to 1. 

Where these lines cross the pocket head line D° F°, 
as indicated by the heavy dots, project these dots of 
intersections horizontally to the right, intersecting sim- 
ilar numbered connected lines in the half rear view (Z), 
as indicated by the intersection 91, 29, 93, 38, 84, 47 
and E. In tins case the corner of the pocket F° in the 
side view intersects the line 5 7 and this corner is pro- 
jected to the line 5 7 in the half rear view, thus obtain- 
ing E. 

The bottom of the pocket F" F° in the side view, 
cuts the line 7 4 at 74. At pleasure establish another 
point on the bottom line of the pocket between the 
center line 14 6 and intersection 74, as shown by a", 
and draw a line from 4 through a" intersecting the middle 
or joint hue 14 6 at a'. Project a' back to the outer 
curve in the rear view, as shown by a and draw a line 
from a to 4, cutting the base line of the pocket F x E, at 
a". The intersections a" and 74 in the rear view then 
indicate similar points a" and 74 in the side view. 
Trace a curve through the various intersections thus 
found in (Z) as from 1 to E, then D 38 E F 1 will be 
the pattern for rear head in the pocket on the line D° F°. 

Space the profiles in the half rear view above the 
half plan into the same number of spaces as in the 
similar rear view (Z) as shown by similar number. From 
the intersections 1 to 5 drop vertical lines (not shown) 
cutting similar lines 1 5 in the plan at 2, 3, and 4. In 
a similar manner from intersections 6 to a to 1 in the 
rear view, drop vertical lines (not shown) cutting sim- 
ilar lines in the half plan at 7, a, 8, and 9. Connect 
opposite points by dotted lines both in the rear view and 
plan in the same manner as in the hah rear view (Z). 
Where the bottom line of the pocket F° F° in the side 
view intersects the lines a' 4 and 4 7 at a" and 74 re- 
spectively, project lines upward until they cut the top 
line X D" at a and 74. Using 14 in the front view as 
center, with radii equal to 14 a and 14 74, draw the 
quadrants cutting the side of the pan in plan, also at 
a and 74, from which points draw horizontal lines inter- 
secting similar numbered lines a 4 and 4 7 in plan at 
a x and 74 1 respectively. 

From the intersection of the corner of the pocket 
F° in the side view, with the line 7 5, project this point 



to the plan view by the quadrant struck from the center 
14 and intersect the similar numbered line 7 5 in the 
half plan at E°. From the intersection b in the rear 
view above the plan, between the bottom of the pocket 
and curve 6 D, project this point to the proper sectional 
line in plan as indicated by b on the line 6 14. Through 
points of intersections thus obtained trace the eurve 
E° to b in plan, which represents the intersects -n or 
miter line between the rear half of the side pocket with 
the rear part of the pan. 

To" obtain the pattern for the bottom and front 
of the pocket on the rear part of the pan, project a line 
from the center c from which the curve al e in the half 
rear view was struck, until it cuts the pocket in plan 
at d° and 9. Take a tracing of E° d° 9 b and place it 
as shown by E° d° 9 b in diagram Y. Erect the lines 
E° d° and b 9 indefinitely, and place on both lines, 
starting from 9 and d°, the stretchout of d e D in the 
rear view, as shown by 9 e° D° and d° e x D 1 , which 
completes the rear part of the pattern for the pocket. 

To obtain the pattern for the transition piece, or 
the rear part of the dust pan, shown by D" 5 6 14 in 
the side view, a set of triangles must be constructed 
showing the true lengths of the various dotted lines in 
either side view of half rear view (Z), and are obtained 
as follows: 

Extend the lines 5 D" and 6 14 in the side view 
indefinitely, as shown by 5 W and 6 V. Starting from 
any point as 5 on the line U W, place the distances of 
the various lines in the rear view (Z) as 5 to 7, 7 to 4 
4 to a, 4 to 8, 8 to 3, 3 to 9, 9 to 2, and 9 to 1, as shown 
by similar numbers or divisions on U W. From the 
numbers 7, 8, and 9 and the letter a erect lines at right 
angles to U W, intersecting the line 6 V at 7°, 8°, 9°, 
and a°. Draw solid lines from 5 to 7°, 7° to 4, 4 to a° , 
4 to 8°, 8° to 3, 3 to 9°, 9° to 2, and 9° to 1, which will 
represent the true lengths of similar numbered lines shown 
in the four views. 

The pattern is now in order and is developed as 
follows: As 5 6 in the side view shows its true length, 
take this distance and place it in diagram T, as shown. 
With 6 7 in the rear view (Z) as radius and 6 in T in 
center, describe the arc 7, which intersect by an arc 
struck from 5 as center and with 5 7° in the true lengths 
as radius. With 5 4 in (Z) as radius, and 5 in T as 
center describe the arc 4, which intersect by an arc 
struck from 7 as center and 7° 4 in the true lengths 
as radius. Proceed in this manner, using first the di- 
visions on the outer profile 6 D in (Z), then the proper 
true length; the divisions on the inner profile 5 D in 



Pattern Problems. 



423 



(Z) and then again the proper true length, until the 
line 1 1 in the pattern T has been obtained, which 
equals the length of 1 1 in the side view. 

From points 1 and 1 in T, at right angles, draw 
the line IDC, on each side equal to the stretchout 
of IDC in the rear view (Z), allowing an edge for 
wiring as indicated by the dotted line in T. Trace a 
line through points thus obtained and C 6 5 C will be 
the pattern shape. 

It will be necessary to find the outline of the shape 
to be cut out of the pattern T, to admit the joining of 
the pocket, as shown by the shaded part D D° E b, 
which is obtained as follows: Where the bottom of the 
pocket F° F" in the side view intersects the lines 
4 a' and 4 7, at a" and 74 respectively, project lines 
upward and intersect similar lines 4o° and 4 7° in the 
true lengths as indicated by a x and 74°. In a similar 
manner where the side of the pocket D° F° in side view 
intersects the various lines, project this line F° D° ver- 
tically, and intersect similar numbered true lengths 
as shown. For example, the line or side D° F° inter- 
sects 9 1, 9 2, and 9 3 in the side view; these similar 
intersections are shown in the true length by 91°, 29°, 
and 93°, and in this way are the balance of the inter- 
sections 38°, 84°, 4a°, 47°, and E° found and corre- 
spond to the numbers shown along the intersecting line 
D E in the rear view (Z) . 

The same intersections on the true lengths can be 
proven as follows: For an example, to prove the inter- 
section 84° in the true lengths: Take the distance 4 84 
in the rear view (Z) on the line 4 8 and place it from 
4 to 84 between 4 and 8 on the line U W. From the 
point 84 erect a perpendicular line to U W, intersecting 
the true length 8° 4 at 84°, and it will be found that 
it meets a similar intersection previously obtained from 
the plan. 

Obtain the outline of the opening in the pattern 
by taking the distance from 1 to 11 in the side view and 
placing it in the pattern T from D to D° and 1 to 11. 
Take the distances in the true lengths, as 1 to 91°, 2 
to 29°, 3 to 93°, etc., and place them on similar lines 
in T, as 1 to 91, 2 to 29, 3 to 93, etc., until the point 
E is placed. Then again take the distances 7° to 74° 
and a° to a x in the true lengths, and place them in T 
from 7 to 74 and a to of . Finally take the division a 
to b in the rear view (Z) and place it in T from a to b 
between a and 8, and trace a line through points of in- 
tersections shown. Then will D° E b be the desired 
outline to be cut out to admit the joining of the pocket. 

The method of obtaining the pocket intersection 



with the front part of the dust pan and the necessary 
patterns will be briefly described and are shown in Fig. 
746. A reproduction of the one half front view, the 
half plan of same, including the side view of the front 
part, has been reproduced in their proper positions as 
indicated in Fig. 746, to the right of the center line 



One ho'f Pattern for Front of Dust Pan 




Full Pattern for Bottom and Front 





of Pocket 


D" 


V" 


1 

I 
1 


1 

1 
1 
1 
1 

1 
1 
1 


D° 


/ 





Pattern for Front \™ 
Head m Poqket 

5k 



1 7rLrp4,pnytHqf5inji!ar 

^nurribertdl/tye^phown 

ynphn, frorJ and $ide 

\! x ! \i 




B One half Front Vie 



Side Mew of Front Part 



Fig. 746. — Half Plan, Half Front and. Side Views, True Lengths 
and Pattern Shape. 



A B. Thus 1 5 8 in the half front view shows the sec- 
tion on the line 1 8 in the half plan and the line 1 8 in 
the side view, while 8 10 12 shows the section on 8 12 
and also on the line 8' 12 in the side view. The side 
view is projected from the plan the quadrants struck 
from the center 8, all as shown. In the front view 
8 Yd shows the front head of the side pocket, indi- 
cated in plan by d a x h° and in the side view by 
h a d° . 



424 



The New Metal Worker Pattern Book. 



The first step is to divide the curves in the front 
view into equal spaces, as shown from 2 to 4, 6 to 8, 
and from 8 to 12. Connect opposite points by lines 
as shown from 12 to 2, 11 to 2 and 3; 10 to 3, 4, 5, 
and 6; 6 to 9, and 9 to 7 and 8. From the various 
intersections 1 to 8 in the front view draw horizontal 
fines (not shown) intersecting the side view from 1 to 8. 
From similar intersections in the front view, erect ver- 
tical fines (not shown) cutting the plan view also from 
1 to 8. In a similar manner, from the points 8 to 12 
in the half front view, draw horizontal and vertical 
fines (not shown), the former intersecting the side view 
from 8' to 12 and the latter intersecting the plan view 
also from 8 to 12. Connect opposite points in both 
side and plan view in exactly the same manner as was 
done in the half front view. 

Complete the side view of the pocket as shown 
by h a d° so that the intersection between the corner 
a of the pocket in the side view, and the front part of 
the pan may be found in the front view, as indicated 
by a°. Draw a line from 11 in the side view through 
the corner a cutting the line 1 8 at a'. Establish at 
pleasure the point b midway between a and d° and from 
a' draw a line through b intersecting the joint line 8' 12 
at b'. In a similar manner, between 9 and 10 in the 
side view, establish at pleasure an extra point c, and 
draw a line from c to 6. 

Project these three points back to the front view, 
as shown by c, b, and a respectively, and draw lines 
from c to 6, and b to a. In a similar manner project 
these points in the plan and draw connecting lines 
as shown. Where the head of the pocket h a in the 
side view intersects the various lines as shown b}' the 
small dots, project horizontal lines (not shown) back 
to the front view intersecting similar numbered lines 
as indicated by 89, 79, 69, 6c, 610, 510, 410, 310, 
a b, a° and b° . 

For example, where the line h a in the side view 
intersects the fine 5 6 10 at w draw the horizontal line 
as shown intersecting the lines 5 10 and 6 10 in the 
front view, and 510 and 610 respectively. Trace a line 
through intersecting points thus obtained, and 8 510 a° 
d Y will be the pattern for the front head 'in the 
pocket. 

From the points a°, b°, and d in the front view, 
erect lines cutting similar lines in plan at a x , b x , and d 
respectively. Project h in the side view to h° in the 
plan view, and complete the plan of the front part of 
the pocket as indicated by h° a x d. From v, the center 
with which the curved corner of the pocket in the front 



view was struck, erect a line in the plan as shown, cutting 
the plan of the pocket at t and r. 

Take a tracing of the pocket pattern Y in Fig. 745 
and place it in Fig. 746 as shown by E° D 1 D° d', 
and to the line d' D° add a tracing of d a x t r in plan, 
as shown by similar letters in the full pattern for pocket. 
Extend D X D° and a x t' until they meet at D°. Then 
E° D x D" a x d' will be the full pattern for the bottom 
and front of the side pocket. 

The true lengths of the lines shown in the front 
view are obtained by extending the lines 12 8' and 1 8 
in the side view, as shown by H J and 8 L, and on the 
line H J the various distances of the fines in the front 
view are placed, as indicated by similar numbers on 
H J. From the various points 2, 3, a, 4, 5, 6, 7, and 8 
at right angles to H J, lines are erected, meeting the 
line 8 L at 2°, 3°, a , 4°, 5°, 6°, 7°, and 8°. Draw 
the true lengths or slant lines to their proper number, 
as shown to correspond 'to similar numbered connections 
in the front view. 

The one half pattern for front of dust pan is now 
in order. Take the true length of 1 12 in the side 
view, and place it as shown by 1 12 in R. With 1 2 
in the front view as radius, and 1 in R as center, de- 
scribe the arc 2, which intersect by an arc, struck from 
12 as center and 12 2° in the true lengths S as radius. 
Then using 12 11 in the front view as radius, and 12 
in R as center, describe the arc 11, which intersect by 
an arc struck from 2 as center and 2° 11 in the true 
lengths S as radius. Proceed in this manner, using 
alternately, first the divisions along the profile 1 8 in 
the front view, then the proper length in S; the divi- 
sions along the profile 8 12 in the front view, then again 
the true lengths in S, until the line 8 8' in the pattern 
R has been obtained, which is equal to 8' 8 in the side 
view. To this line 8 8' add the girth of the flange and 
wire edge 8 W in the front view, as indicated by 8' W 
in R. 

As the reference letters and figures are similar in 
the pattern, true lengths and front view, the opera- 
tions just given, and which will follow, can be easily 
followed. To obtain the intersecting points on the true 
lengths, between the pocket and pan, simply extend the 
side a h and extend a vertical line from b in the side 
view, cutting similar numbered lines in the true lengths 
as shown. These points of intersections have been 
further proved by taking the various divisions where 
the miter line intersects the various lines in the front 
view, and placing them between their proper numbers 
on the line H J, and where by erecting perpendiculars 



Pattern Problems. 



425 



to H J similar intersecting points are met, as previously 
obtained from the side view. 

To obtain the opening to be cut out in the pattern 
R to admit the joining of the side pocket, take the dis- 
tance from 8' to h in the side view and place it from 
8' to h in E. Then take the distance from 9 to 89° 
in S and place it from 9 to 89 in R. Proceed in this 
manner, taking the true lengths from S and placing 
them on similar lines in R until the point 6° has been 
obtained. 



Take the distance from b to d in the front view 
and place it in the pattern R between the points b and 
10, as shown by d. Trace a line through points thus 
obtained, as shown shaded, which will be the desired 
opening. 

The intersecting points in the pattern are numbered 
similar to those in the true lengths S so that the proDer 
line from which the distance was obtained can be located. 

Edges and locks must be allowed on all patterns 
for either riveting or seaming. 



PROBLEM 220. 



Making a Three-prong Fork in Two Pieces. 



A quick and novel method of laying out the work 
so that only three seams are required and no double seam- 
ing is necessary. It has been found that the fork can 
be made out of less material and in a great deal less 
time than with methods commonly employed. It is 
made in two pieces and formed entirely in the brake 
with seams as indicated. When finished, the fork looks 
well and is perfectly smooth inside and outside. This 



Elevation 
.-Collar Seam /Co/for Seam .Collar 




Plan 



Fig. 7 47. — Plan and Elevation of Three-prong Fork. 

method is adaptable to any profile and any angle. The 
first fork used was to convey water and it has been 
tried with success on general piping work. 

The fork as shown in plan and elevation in Fig. 
747 gives a novel method of construction, thereby mak- 
ing the development of the pattern much quicker than 
the customary fork in use. The development of the 
patterns is given in detail in Fig. 748 in which but one 
half elevation is shown, as both ends are similar. 



On the different ends of the openings in the prong, 
place the profiles as shown. Thus, first using 1' as a 
center, draw the quarter circle 4' 1. Using 8' as center, 



Half Elevation 




Alfow Laps for 
Riveting 



Fig. 74S. — Method of Developing Patterns for Three-prong Fork. 

draw the semi-circle 5' 11' and using 15 as center, draw 
the quarter circle 12' 15. Divide these sections into an 
equal number of spaces, as shown respectively from 



426 



The New Metal Worker Pattern Book. 



1 to 4', 5' to '11, and 12' to 15, being careful that the 
number of spaces between 8' and 11 are similar to the 
spaces between 12' and 15. At right angles to 1' 11' 
in elevation and from the various, intersections 1 to 3 
and 6 to 10 in the sections, draw lines intersecting the 
base line 1' 11', as shown by similar numbers. In a 
similar manner from the intersections 13 and 14 in the 
quarter section, draw lines at right angles to 12' 15', 
intersecting the line 12' 15' at 13' and 14'. Draw lines 
from points 1' to 4', to 15'; from points 5' to 8', to 
15'; and from 8' to 14' to 9' to 13' to 10' to 12'. These 
lines represent the bases of sections which will be con- 
structed in diagrams L, M and N, the altitudes of which 
will equal the various similar numbered bights in the 
sections in elevation. 

To obtain the true lengths from 15' to 1', 2', 3', 
and 4', place them on the line D E in diagram L as 
shown by similar numbers, from which points erect 
perpendiculars equal to similar numbers in the sections 
in elevation. For example, to find the true length of 
15' 2' in elevation, take this distance and place it on the 
fine D E in diagram L and make the distances 15' 15 and 
2' 2 equal to similar distances in the sections in elevation. 
A line drawn from 2 to 15 in L is the true length of the 
fine 2' 15' in elevation. In a similar manner take the 
various lengths of the lines in B and C in elevation and 
place them on the lines F G and H J respectively, as 
shown by similar numbers, and obtain the vertical nights 
from similar numbers in the sections in elevation, all 
as shown by similar numbers in both diagrams M and 
N. The true lengths having been obtained the pattern 
is developed as follows: Take the true length 1 15 in : 
diagram L and place it on the vertical line 1 15 in the 
pattern P. Now with radii equal to 15 2, 3 and 4 in 
diagram L, and 15 in P as center draw short arcs 2, 
3, and 4. Set the dividers equal to the spaces 1 to 4 



in the section in elevation, and starting from 1 in P, 
step to arcs 2, 3, and 4, and draw a fine from 4' to 15. 
With a radius equal to 4 5 in elevation and 4 in P as 
center, describe the arc 5 which intersect by an arc 
struck from 15 as center and 15 5' in diagram M as 
radius. Draw lines in P, from 4' to 5 to 15. With 
radii equal to 15 6, 15 7, and 15 8 in the true lengths 
in M and 15 in the pattern P as center, describe the short 
arcs 6, 7, and 8. Set the dividers equal to the divi- 
sions between 5' and 8 in the semi-section in elevation, 
and starting from 5 in P, step to arcs 6, 7, and 8 as 
shown. With 15 14' in the section in elevation as radius, 
and 15 in P as center, describe the arc 14 which inter- 
sect by an arc struck from 8 as center, and the true 
length 8 14 in diagram N as radius. Now, with 8 9 
in section in elevation as radius and 8 in P as center, 
describe the arc 9, which intersect by an arc struck 
from 14 as center and 14 9 in diagram N as radius. Pro- 
ceed in this manner, using alternately first the division 
in the lower profile in elevation, then the true length 
in N; the proper division in the upper profile in ele- 
vation, then the true length in N, until the line 11' 12' 
in the true length in the pattern has been obtained, 
which is equal to 11' 12' in elevation. Trace the out- 
line in the pattern P as shown by 1 4' 5'11' 12' 15 and 
trace this hah pattern opposite the line 1 15 as indi- 
cated by 14 5 11 12 15, which completes the half pat- 
tern for the three-pronged fork. 

Laps for riveting and seaming on the collars should 
be allowed for. Then cut out the metal and it is ready 
to be taken to the brake and any workman who is familiar 
with the use of the brake can soon form up the two 
parts of the fork so that it is an easy matter to rivet them 
together, at the first trial at making one, into a fork that 
will have a decidedly pleasing appearance especially when 
the expeditious procedure is considered. 



PROBLEM 221. 



A Spiral Conveyor or Loading Spout. 



The problem of Fig. 749 assumes that the loading 
spout is to have a pitch of 45 degrees with a reverse 
pitch on the same angle, the template for the spout to 
be semi-elliptical and 14X32 inches in size. Drawings 
have been prepared, making the angle of the spout in 
elevation 60 degrees in place of 45 degrees, so as to show 
more clearly the principles involved. In this case the 



pitch in the plan is assumed to be as much as indicated 
by B C. In this connection it is proper to say that no 
matter what the angle may be in the elevation or the 
amount of offset in plan, the principles hereinafter ex- 
plained will apply to any angle, offset or section. 

Establish at pleasure any point A, from which draw 
the horizontal line A 9 and the vertical line A B, forming 



Pattern Problems. 



427 



a right angle as shown in Fig. 750. From A draw an 
angle of 60 degrees as shown by A c. Establish at pleasure 
the throat in elevation as A 1 and using A as center 
and A 1 as radius, describe the arc 1 b. Make the dis- 
tance 1 9 equal to 32 inches, as called for, and using 



draw radial lines to the center A, cutting the inner are 
as shown. Extend these radial lines until they inter- 
sect the outer arc. 

These radial lines then become the miter or joint 
lines, thus allowing a given profile on the line 1 9 in 



--14'-. 



"Elevation 




':!? P 



5 6 



7 &9 

' ! TrueLengfhsof 
5olid Lines 



It.9 ?345'67'8'9 
True Lengths of 
Doffed Lines 




Fig. 750. — Method of Developing Pattern for Loading Spout. 



A as center describe the outer arc 9 c. Bisect the line 1 9 
in elevation and obtain the point 5. Using A as center 
and A 5 as radius, describe the arc 5 a. As the spout 
is to have four pieces, divide the arc 5 a into four equal 
parts, as indicated by 5, 5°, e, d, a, through which points 



elevation, to become the true profile on each of the 
joint lines shown. Thus it will be seen that by dividing 
the elevation into equal parts, each piece and each joint 
line becomes equal in length, so that the same profile and 
girth can be used on every one of the radial fines shown. 



428 



The New Metal Worker Pattern Book. 



By using this method and that which will follow, 
only one pattern need be developed, which can be used 
for the four pieces, no matter what the offset plan may be. 

From any point B on the center line A B, draw 
the horizontal line B 9 and intersect it by a line dropped 
from 5 in elevation at 5 Z in plan. This point 5 J , then, 
represents the center, around which the profile shown 
by 1 5 9 is placed in position and represents the true 
profile on the line 1 9 in elevation. Knowing the amount 
of offset that the spout is to have, set off this distance 
as shown from B to C and divide this distance into as 
many parts as the spout will have pieces (in this case 
four), as shown by F, E, D, and C. From the points 
C, D, E, and F, draw lines parallel to B 5 X , and inter- 
sect them by vertical lines (partly shown) dropped from 
points a, d, e, and 5° in elevation, at a°, d°, e°, and 5" 
respectively, in plan. Draw a line from a° to d° to e° 
to 5" to 5 X , which will represent the center line of the 
spout in plan and will show how far each one of the 
four pieces in elevation will project away from the reader 
the first piece starting at 5 J 5°, and the fourth piece 
at d° a°. 

In working out this problem in practice, all that 
is required so far as the pattern is concerned, is the 
center line 5 X 5" in the plan view of the first piece in 
elevation. 

As the profile 15 9 in plan shows the true section 
on the line 1 9 in elevation, the next step is to obtain a 
horizontal section on the line 1° 9° in elevation, as fol- 
lows: Divide the true profile in plan into equal spaces 
as shown by the small figures 1 to 9. From these 
divisions erect vertical lines (not shown), cutting the 
base line of the first piece in elevation as shown 
from 1 to 9. Using A as center, with the various divi- 
sions 1 to 9 as radii, describe arcs cutting the radial 
line or joint line of the first piece, as shown from 1° 
to 9°. Extend the line F 5° in plan as shown, on which 
place a reproduction of the true profile with the various 
intersections, as shown by Y. From the various di- 
visions in Y draw horizontal lines to the left, which 
intersect by vertical lines dropped from similar num- 
bers on the joint line 1° 9° in elevation, thus obtaining 
the intersections 1° to 9° in plan. Trace a line through 
points thus obtained and 1° to 5° 9° will be the hori- 
zontal section on the joint line 1° 9° in elevation. 

Connect solid and dotted lines in plan as shown 
and connect similar numbers in a similar manner in the 
elevation. Then will these solid and dotted lines in 
plan represent the bases of triangles which are yet to 
be constructed, the altitudes of which will equal the 



vertical hights of similar numbered intersections in 
elevation. 

To obtain the length of the solid lines, proceed as 
follows: Extend the line A 9 in elevation as shown and 
from any points on same, as a a and b b, erect perpen- 
diculars, which intersect by horizontal lines drawn from 
the various intersections on the joint line 1° 9° in ele- 
vation, thus obtaining the points 1° to 9° on lines erected 
from a a and b b. 

As all the solid lines in plan are similar, take the 
distance of one and place it to the right of the eleva- 
tion, from the corner a a to the point marked 1 to 9, 
and from this point draw lines to the various inter- 
sections 1° to 9°, which will give the true lengths of 
the solid lines in plan or elevation. 

Obtain the true lengths of the dotted lines by taking 
the various lengths of the dotted lines in plan, and 
setting them off from the corner b b in the diagram of 
clotted lines as follows: Take the distance from 8° to 
9 in plan and set it off from b b in the diagram and mark 
the intersection 9. As the dotted line runs from 9 to 
8° in plan, draw a line from the point 9 just obtained 
in the diagram to the point 8° on the perpendicular 
line erected from b b. Then 8° 9 is then the true length 
of 8° 9 in plan or elevation. In a similar manner take 
the various distances in plan from 7° to 8, 6° to 7, 5° 
to 6, 4° to 5, 3° to 4, 2° to 3, and 1° to 2, and place 
them in the diagram of dotted lines, measuring in each 
instance from the point b b and draw lines from 7° on 
the perpendicular line to 8 on the base line; 6° to 7, 
5° to 6, 4° to 5, 3° to 4, 2° to 3, and 1° to 2, which rep- 
resent the true lengths of the dotted lines in plan or 
elevation. 

The pattern is now in order and is developed as 
follows: Draw any horizontal line 1 1° in X, equal to the 
true length 11° in R. With 1 2 in the true profile 
as radius and 1 in X as center describe the arc 2, which 
intersect by an arc struck from 1° as a center and 1° 2 
in diagram of dotted line T as radius. Using the same 
radius 1 2 of the true profile with 1° in X as center, 
describe the arc 2°, which intersect by an arc struck 
from 2 as center and 2 to 2° in diagram R as radius. 
Proceed in this manner, using alternately first the proper 
division in the true profile, then the proper solid line 
in diagram R; again the proper division in the true 
profile, then the proper line in diagram T, until the 
line 9 9° in the pattern has been obtained. Trace a 
line through points thus obtained, then will 1 9 9° 1° 
be the pattern shape for all four pieces for the spout., 
to which riveting flanges must be allowed. 






Pattern Problems. 



429 



PROBLEM 222. 



Developing Patterns for an Unusual Two-way Y. 



In this problem both legs of the two-way Y are to 
be of equal but short length ; the leg A, however, is to be of 
a different angle from the leg B, the two legs to form 
an acute angle in plan. The opening of the feed collar 
to be square to the plumb line. 

In this connection, it is only proper to say that it 
makes no difference how long the legs may be, whether 
of equal length or not, or what diameters the three 



Tru_ _ 
of Left 




Full Pattern 
■for Left Prong 2- 



Fig. 751. — Method of Developing Patterns for Two-way Y. 

openings may have or what angles the legs may have 
in plan or elevation, the principles set forth herein are 
applicable to any condition. 

The first step is to draw the correct angle of the 
two branches in plan as indicated by C 4 E, and using 
4 as center with the required radius, draw the plan of 
the feed collar shown by 1, 11, 6, 8. Now extend the 
lines C 4 and E 4 until they intersect the circle at D 
and F respectively. As both branches have different 
angles in elevation, true elevations must be drawn at 



right angles to C 4 and E 4 respectively. Bisect the 
angle C 4 E in plan and draw the miter or joint line 
A B, cutting the circle at point 6. At right angles to 
4 E and 4 C tangent to the circle, draw lines inter- 
secting each other at 2. Then 6 2 is the length of the 
joint line in plan, through which a true section or profile 
must be drawn at pleasure as follows: Parallel 2 6 in 
plan, draw any line as 1 6 in M, at right angles to 
which, from points 2 and 6 in plan, erect lines inter- 
secting the line 1 6 in M as shown. Now take one-half 
the diameter of the circle R in plan and set it on the 
center line in M, from / to 4 and draw at pleasure the 
graceful semi-elliptical figure shown by 1 2 4 6. Divide 
this true section M in equal parts, making the spaces 
smaller where the curve is shorter as shown. From these 
divisions numbered 1 to 6, at right angles to A 6 in 
plan draw lines intersecting A 6, also from 1 to 6. As 
C 4 in plan represents the center line of the left prong, 
a true elevation must be found on this line; also a fore- 
shortened view on the joint line 2 G in plan, and is ac- 
complished as follows: From 4 in plan extend the center 
line 4 G, cutting any horizontal line drawn at pleasure, 
as 6' 10', at a. From a draw the angle G a 15, as de- 
sired, making the line a 15 as long as required. Through 
15, perpendicular to 15 a, draw the diameter, 13 17 
of the required size. Draw a line from 13 to 10'. If 
13 10' is required to be longer or shorter, this can be 
made as desired, which will simply change the length 
of the center line 15 a. At right angles to 13 17 draw 
the profile of the opening on 13 17, as shown by N, 
which divide into equal spaces shown from 13 to 20, 
from which points perpendiculars are drawn to 13 17 
until they intersect 13 17 in elevation as shown by 
similar numbers. Now from the intersection 1 to 6 . 
on the joint fine 2 6 in plan, erect perpendicular lines 
parallel to 4 G, until they cut the line 10' 6', which 
represents the diameter of the feed collar, as shown. 
Now measuring from the line 1 6 in the true section M, 
take the various distances to points 2 to 5 in the section, 
and place them in the elevation of the left prong on similar 
fines just erected, measuring in each instance from the 
fine 6' 10', and resulting in a foreshortened view of the 
joint fine between the two prongs, as indicated by 
similar numbers. Draw a fine from 17 tangent to this 
view at 5', which completes the elevation of the left 
prong. 



430 



The New Metal Worker Pattern Book. 



A plan view must now be obtained on 13 17 in 
elevation, as follows: From the various intersections 
13 to 17 in elevation, drop vertical lines crossing the 
center line C 4 in plan, as shown. Now, measuring 
from the line 13 17 in the profile N, take the various 
divisions to the various numbers 14 to 20 and place 
them on lines having similar numbers, measuring on 
each side of the center fine C 4 in plan, and resulting 
in the elliptical figures shown by 13, 15, 17, 19. In 
a similar manner obtain the elevation of the right prong. 
At right angles to E F in plan construct the true ele- 
vation as before explained in connection with the left 
prong. It will be noticed that the angle H 6 27 is 
different from the angle G a 15 in the left prong. The 
fine 21 25 in the right prong is drawn at right angles 
to b 27, and the distance 21 9" is made equal to 13 10' 
in the left prong, although 21 9" in the right prong 
could be made any desired length. The foreshortened 
view 1" 4" 6" in the right prong, as well as the plan 
view on the fine 21 25, is obtained as previously de- 
scribed in connection with the left prong. In this case 
the profile T is similar to the profile N. It makes no 
difference whether these two profiles are similar or not, 
as the principles can be applied to various measure- 
ments and sizes. 

Having drawn the two true elevations of the prongs 
at right angles to their respective center lines in plan 
as well as the plan of the prongs and feed collar, fines 
must now be drawn in plan representing the base fines 
of triangles, which must be. constructed. All that por- 
tion of the feed collar between 1, 11, and 6 will be 
connected to that part of the outlet between 13, 15, 
and 17; while the portion between the miter joint 1, 
2, 4, 6 in plan will be connected to the opposite side 
of the outlet, as indicated between 13, 19, 17. Thus 
the points are connected as follows: 1 to 13 to 10 to 14 
to 11 to 15 to 12 to 16 to 6. The lower half is con- 
nected from 17 to 6, 5 and 4 to 18; then 18 to 3 and 2; 
then 2 to 19 and 20 to 1. These various lines in plan 
then represent the base fines of triangles which will be 
drawn, whose altitudes are equal to the horizontal dis- 
tances between equal numbered points in the left elevation, 
as shown parallel to 10' 6'. 

For example, to find the true length of the line 
10 14 in plan, take this distance and set it off on the 
fine 10' 6° in the elevation of the left prong, as shown 
by 10 14. From point 14 drop a perpendicular fine until 
it intersects the horizontal line drawn from point 14 in 
the elevation at 14°. Draw a line from 14° to 10, which 
is the true length of the line shown by 14 10 in plan. 



It will be noticed that part of the true lengths 
have been placed below P, for want of space. In this 
diagram the vertical distances are all obtained between 
similar numbered points in the left elevation. For 
example: The true length of the fine 17 to 6 in plan is 
obtained by placing this distance as shown from 17 
to 6 in P, and from 17 a perpendicular line is drawn, 




Riqht'Pronq 
y"i Elevotidn 



37' r 

Fig. 752. — View of Y Developed. 

making 17, 17° equal to the distance between the 
horizontal lines drawn from 6 and 17 in the left eleva- 
tion. The line drawn from 17° to 6 is the desired true 
length. 

In obtaining the true lengths of the right prong, 
the fines are connected in similar manner as shown 
in plan, the altitudes being equal to the various num- 
bered divisions between c and d in the right prong. 

The full pattern for the left prong is developed as 
follows: Assuming that the seam is to come along 13 1 
in plan, take the length of 13° 1 in the true lengths in 
P and place it as shown by 13 1 in S. Now using 1 10 
in the circle R in plan as radius and 1 in S as center, 
describe the arc 10, which intersect by an arc struck 
from 13 as center and 13° 10 in P as radius. Now, 
with 13 14 in the profile N as radius and 13 in S as 
center, draw the arc 14, which intersect by an arc struck 
from 10 as center and 10 14° in P as radius. Proceed 
in this manner, using alternately first the division be- 
tween 1116 in R, then the proper length in diagram 
P; the division in the profile N, then again the proper 
length in P until the line 6 17 in S has been drawn, 
after which the balance of the divisions between 16 and 
13° are obtained from similar numbers in N, and the 
divisions from 6 to 10° in S are obtained from the true 
section M, and the proper slant lines in P are placed in 
their proper position in S. 



Pattern Problems. 



431 



Note that points 5 and 4 in S are drawn to 17; 
4 3 and 2 to 18; 19 and 20 to 2, etc. Whichever way 
the lines are connected in plan, so must the lines con- 
nect in the pattern as shown. Trace a line through 
points of intersections thus obtained as shown by 



1 6 1° 13° 13, which will be the full pattern for the left 
prong. 

After the true lengths of the lines have been found 
for the right prong, the pattern is developed the same 
as just described. 



PROBLEM 223. 
Pattern for Elbow Transition Piece. 



This problem demonstrates the shortest method 
of developing the patterns for an elbow, measuring 
10X48 inches at one end and making a transition to 
a 13-inch round pipe at the other end. The radius 
of the throat and heel of the elbow is shown in Fig. 
753. In this figure the section on the line A B is a 
rectangle, 10X48 inches. The heel of the elbow is 



., P <? 




i 


Cw '< 

V J\\ ;--,e.1oi 

* f D \\ 
Section _; 
through ©, 

WtS'- 


&* 


"\ 


a 

Sic 


32'rad. 

B 

e Elevation 



Section through A-B 



■48 



Fig. 75S.—The Problem. 

struck with a 32-inch radius with a as a center and the 
throat is struck from b with a 12-inch radius. The 
section on the line C D is a true circle, 13 inches in 
diameter. 

To lay out this elbow in sections so that a true 
taper may be obtained from the 10-inch sides in the 
rectangle to the points x and x in the 13-inch pipe, would 
require more time and labor than the estimated cost. 
In work of this kind, as often made, short methods for 
obtaining the same results can be used, providing, how- 
ever, that the area is not decreased at any point. There- 
fore, in making up the elbow, draw a joint line from 
c to d or the points of tangency between the straight 
lines of the round pipe and the throat and heel of the 
elbow, and cut two sides as shown by c d B A, to which 
is doubled seamed the 10-inch strip representing the width 
of the duct or elbow. Thus it will be seen that the 
elbow from c to A and B to d will be 10 inches wide 



throughout, while the transition will be made between 
the rectangular joint line, c d, and the round pipe e f. 

The proper way to develop the pattern for this 
transition piece is shown in detail in Fig. 754, in which 
5 3' 2' 11 is an enlarged reproduction of e d cf in Fig. 
753. On 5 11 draw the half section of the 13-inch 



- IcyeLsnaths in B . 



6 

f -5zer7- - - ■ 



— Wb= 




Elevation of Transition 
Piece with Semi-Sections '*** 



True Lengths in A. 




Reverse on th/'s L/ne _ 

One-Half Pattern Shape 
A] low Edges for Seaming 
oea/n 2 

Fig. 754- — Details of Pattern for Transition Piece. 



pipe, and on 2' 3' the half section of the rectangular 
joint line, the distances 2' 2 and 3' 3 being 5 inches, 
or half the distance of the 10-inch ends drawn at right 
angles to 2' 3'. 

The hah profile of the round pipe is next divided 
into equal parts as shown by the small figures 5 to 
11, from which lines are drawn at right angles to 5 11, 
intersecting this line from 5 to 8' to 11, as shown. From 
6', 7' and 8' lines are drawn to the corner 3', while 
from 8', 9' and 10' lines are drawn to the corner 2'. 
These lines then represent the bases of triangles which 



432 



The New Metal Worker Pattern Book. 



will be constructed with altitudes equal to the various 
hights in the semi-section. Thus, the true lengths 
of the lines shown in B are shown above the elevation, 
while the true lengths in A are shown below to the right. 

For example, to find the true length of the line 
8' 2', in A take this distance and place N on the horizontal 
line in K, as shown from 8' to 2', at right angles to which 
erect perpendicular lines from 8' and 2', making 8' 8 
and 2' 2 equal respectively to 8' 8 and 2' 2 in the half 
sections in elevation. A fine drawn from 8 to 2 in K 
will be the true length of the fine 8' 2' in the elevation. 
In this manner all of the true lengths are obtained, as 
shown by similar numbers in the diagrams J and K. 

The pattern is developed as follows: As 5 3' in 
elevations shows the greatest true length for the transi- 
tion piece place this distance on any line as shown 
by 5 3' in diagram L. At right angles to this line draw 
a perpendicular from the point 3', as shown by 3' 3 
equal to 3' 3 in the half section. Draw a line from 3 
to 5 in L. Now, using as radii the true lengths, 3 6, 
3 7, and 3 8 in diagram L, with 3 as center in diagram 



L, draw the arcs 6, 7, and 8. Set the dividers equal 
to 5 6, 6 7, and 7 8 in the half section, and starting 
from 5 in the pattern, step to arc 5, then to 7 and 8, 
and draw a line from 8 to 3. With 3 2 or 48 inches in 
the half section as radius, and 3 in L as center, describe 
the arc 2, which intersect by an arc struck from 8 as 
center and the true length 8 2 in diagram K as center. 
Draw a line from 3 to 2 to 8 in L. 

Using as radii the true lengths 2 9, 2 10, and 2 11 
in diagram K, with 2 in L as center, draw the arcs 9, 
10, and 11. Set the dividers equal to the spaces between 
8 and 11 in the half section and starting from 8 in dia- 
gram L, intersect similar arcs as shown. Draw a fine 
from 11 to 2. Using 11 2' in the elevation as radius 
and 11 in L as center, draw the arc 2', which intersect 
by an arc struck from 2 as center and 2 2' in the half 
section as radius. Draw a fine from 11 to 2' to 2 and 
trace a line through the points of intersections between 
11 and 5. Then will 3' 2' 11'5 be the half pattern for 
the curved elbow transition piece to which seaming 
edges must be allowed. 



PROBLEM 224. 



Pattern for Spiral Strips in Dust Separator. 



It is to be remembered that any geometrical prob- 
lem involving a helix requires considerable thought 
for its solution. The fact is that a true pattern of a 
helix or spiral cannot be developed. 

Some problems of this nature appear in Vol. 12 of 
"Practical Sheet Metal Work and Demonstrated Pat- 
terns." 

Inasmuch as only an approximate pattern can be 
developed, it seems really immaterial what system is 
followed and by reason of the shape of the spiral, or 
rather, the elements of it, not being parallel or radial, 
triangulation is perhaps about the shortest and best 
method to be followed and is recommended for this 
case. 

Therefore, the following demonstration is based on 
that system: Fig. 755 shows a plan and elevation of 
the cone part of the separator. To develop the spiral 
line representing the edge of the strip touching the in- 
side of the cone, divide the plan into an equal number 
of spaces, as to 11. Draw hnes from these points to 
the center. 



As the spiral is to make two and one-half revolutions 
inside the cone, divide the liight of the cone into two 
and one-half equal spaces, as A B C D. Divide the 
spaces A B and B C into twelve spaces (same as the 
plan). Space C D is divided into six parts, as it rep- 
resents but one-half revolution of the spiral. 

Project points to 11 in plan, to line D H of ele- 
vation, giving points on that line D 9, 8 10, 7 11, 6 0, 
5 1, 4 2, and 3 H. Draw lines from these points to 
apex of cone F. These lines are technically known as 
the elements of the cone. Project horizontal lines from 
points on line A D to intersect proper element. As, 
for instance, D intersects 6 0, then point below D would 
intersect element 7 11. Second point below, element 
8 10, and so on, which will give the spiral fine as shown. 
Drop points of intersection of spiral with elements in 
elevation to elements in plan, like T in elevation, which 
is T° in plan. This gives the spiral in plan. 

Choose the width of strip and set it off on fine 0, 
in plan as O P. Set this space all around the spiral 
(although not necessary for the development of the 



Pattern Problems. 



433 



pattern) and the plan of the strip is complete. If a 
complete elevation of the strip is required, the inner 
edge of it would be projected to the elevation, giving 
the appearance as shown from D to C. This is only 
confusing, however, and would be dispensed with in 
practice and the patterns can now be developed from 
these data. 

Develop pattern of cone by selecting a convenient 
point, as F in Fig. 756, for center and with a radius 



9 8-10 



7-11 



6\0 




Fig. 755. — Plan and Elevation. 

equal to F D of Fig. 755, describe a long arc. Set off 
on this arc the spaces to 11 of the plan of Fig. 755. 
Draw lines from these points to point F. With F of 
Fig. 756 as center, describe an arc of a radius equal to 
F G of Fig. 755. Using F in Fig. 756 as center always, 
set off on proper lines the spaces on line F D of Fig. 
755, as was done with space F G. Like F T of Fig. 
755 is F T of Fig. 756; that is to say, you project points 
on the elements in elevation to line D F like C° is C x 



on line F D, then C in Fig. 756. Always being sure to 
place the point on the right element — C° in Fig. 755, 
is on element 6, therefore is on line 6 F of Fig. 756. 




Fig. 756. — Pattern of Cone. 

Drawing a line through the points so obtained in Fig. 
756 gives a guide line on the cone for laying out the 
rivet holes for the rivets to attach the spiral strip. 



'-■r 

Fig. 757. — Diagram of Dotted Lines. 

In this scheme of developing the pattern of the 
spiral strip, its inner edge line will be ignored for the 
time being and a pattern developed for a spiral strip 
that touches, throughout, the center or axial line of 
the cone. To illustrate, a pattern will be developed of 
part of plan of Fig. 755, F°, T°, S°. and not T°, S°, V°, 
W°. Now, while lines S° F°, and T° F° are shown in 



434 



The New Metal Worker Pattern Book. 



their true length, part S° F° T° is not shown true, be- 
cause line T° F° is below S° F°, as much as Y Z of 
the elevation; so it will be necessary to use a dotted 




Fig. 758. — Pattern of First Half Revolution. 

line to Y T, as in triangulation. The true length of 
this dotted line is found in the customary way (as in 
Fig. 757), by drawing a horizontal line Z and a ver- 





Fig. 759.— Pattern of First 
Full Revolution. 



Fig. 760. — Pattern of Last or 
Bottom Full Revolution. 



tical line Z Y. Space Z Y of Fig. 757, is equal to space 
Z Y in Fig. 755. On the horizontal line Z O of Fig. 
757, set off from Z the distance F° T° of the plan in 
Fig. 755. Then from T° to Y in Fig. 757 is the true 



length of the dotted line Y T in elevation of Fig. 755. 
Do this with all the spaces in the plan of Fig. 755, and 
the diagram of Fig. 757 is the result. 

For the pattern of the spiral, draw a line as F 
in Fig. 758, the same length as F in plan Fig. 755. 
From F draw a short arc, the radius of Y of Fig. 757. 
From of Fig. 758 draw an arc to intersect the one 
just drawn; the radius of the arc to be equal to space 
O J of the plan in Fig. 755. From J of Fig. 758 de- 
scribe an arc equal in radius to line F° J of plan in Fig. 
755. From F in Fig. 758 describe an arc to intersect 
this one, the radius to be equal to space Y Z of eleva- 
tion of Fig. 755. Proceeding like this will give pat- 
tern of the spiral as shown in Figs. 758, 759, and 760. 
These are the patterns for a strip that touches the axial 
line of the cone. 

It is to be understood, of course, that the space 
J of the plan, Fig. 755, is not, strictly speaking, cor- 
rect in length. It was suggested that this space be 
taken because it is somewhat shorter than the space 
actually is. This suggestion was made by reason of 
the strip, in practical work, always growing to the extent 
of being altogether too long over all, which naturally 
would give considerable trouble when fastening it to 
the cone. Of course, the actual length of this space 
can be had by measuring a like space on the spiral out- 
line of the cone pattern. 

The next step will be to cut from this a strip of 
the chosen width. As was done to give the plan of 
the full strip, set off all the width on the lines on the 
pattern to wit : P of Fig. 758 equal to OP of the 
plan in Fig. 755. 

In conclusion it is again stated that this is but 
an approximate pattern. However, as the strip is very 
narrow, and as the inner edge does not fasten nor travel 
on anything, the slight waving or buckling of the strip 
will in no way prevent it from working properly. 



PROBLEM 225. 



Faucet Boss Problem. 



The method of developing a pattern for a faucet 
boss to strengthen the connection of the faucet with the 
cylinder or can is the subject of this discussion. 

In the scheme for developing the side elevation 
has been omitted, only the front elevation and plan 
being reproduced with diagrams of the triangles re- 



quired in getting measurements for laying out the pat- 
tern. A study of the problem will reveal that the object 
structurally consists of two differently shaped parts, a 
half diamond 5 16, and a curved part, closely resembling 
a frustum of a cone, 5 12 6 and 13 to 19 indicated in the 
elevation and requiring separate treatment, as Fig. 761. 



Pattern Problems. 



435 



The curved part of the boss, not being exactly a 
cone, a pattern for it cannot be developed by the radial 
lines method, but it can best be developed by the tri- 
angulation method. The reason it is not a true cone 
is because the design calls for a true circle 5 11 6, when 
viewed in elevation; whereas, if it was a true cone in- 
tersecting a cylinder that line would appear irregular 
A cone to show a true circle must intersect a flat surface 
at right angles to the axis of the cone. Furthermore, a 
section of the diamond on line 5 19 6 is a true triangle, 
whereas a section of a true cone on the line would be 
curved, or, strictly speaking, a hyperbola. Hence, the 
object not showing in elevation, the line of intersection 
of a cone with a cylinder and a section on the line 5 19 6 
not being a hyperbola but a triangle, the object is not 
a true cone, but an irregularly curved body. 

The diamond part could be developed by the parallel 
line method. However, it would require just as much 
labor as by triangulation and, as that method is to be 
used for the curved part, triangulation has been adopted 
for both developments. As in Fig. 762, to develop the 
pattern of the diamond part, proceed as follows: 

Drop the point in elevation to the plan as 5". Divide 
that part of the cylinder into equal parts as 5 X 4 X 3* 2 X 
and l x and project these points to the line 5 1 in eleva- 




Fig. 761. — Flan and Elevations. 

tion, as shown. The distance 5 to 4, and so forth, is 
not exact, as shown in elevation, and as it is necessary 
to know that distance, the next step will be to flatten 
out the curved line 5 X to l x . This has been done to 
the left of the elevation in diagram C by placing the 
space 5 X 4 X , etc., on the line 5' 1", as shown. Verticals 
are erected from these points, which in turn are in- 
tersected by lines projected from 4 to 1, as shown. Con- 
necting these points of intersections 5' 4' 3' 2' 1' gives 
the true length of the line -5 to 1 of the elevation. 



It is now necessary to determine the precise length 
of the lines 1 12, 2 12, etc.; so, continue the line 16* a, 
in plan, to the left and erect a perpendicular line from, 
say, point 19 s . To this line project the points 5 X , 4 X , 
etc., containing the projecting lines indefinitely to the 
left. On these lines place the respective distances of 
the elevation, as, for instance, 4 19 in elevation is 4 s 19° 
in the diagram of triangles; these are left for the time 
being, while the correct lengths of the dotted and solid 
lines of the curved part of the boss are found. 

To avoid mixing the lines, the round part of the 
boss at the right in the elevation, is divided into equal 
parts. These are projected clown to the cylinder in 




Fig. 763.— Pattern for Faucet Boss. 

plan as shown. The circular opening of the boss for 
the faucet, in elevation, is also divided into the same 
number of equal parts as indicated. These various 
points of division are given convenient symbols or num- 
bers for designation. As customary, connect the points, 
with dotted and solid lines. 

Having clone this, project to the left the points 
in the outer curve of the boss, that of the faucet opening 
not being required as the line is of true length, as shown 
in elevation. On the line in diagram D, which is marked 
12' 20 s , place the spaces in the plan 12* 11*, etc., meas- 
uring always from 20° in the plan at the right and from 
20 B in diagram D. Erect verticals and proceed as was 
done in diagram C, which gives the true length of the 
outer curve of the boss. 

The correct lengths of the dotted and solid lines 
are determined by practically the same method as for 



436 



The New Metal Worker Pattern Book. 



the diamond part of the boss. For the want of space, 
however, instead of projecting the spaces on the cyl- 
inder, to the right or left, in plan, they are only pro- 




I9 D 
Section on Line 5-6 



Q /8°°iY^J3 ' 

16°° '15°° 
Diagram of Diagram of 

Dotted Lines Solid Lines 



Fig. 762. — Method of Preparing to Develop Pattern for Faucet Boss 



jected to the center line. Then, elsewhere, the dis- 
tances on the center line, as for example, 13* 7 P are 
placed on vertical lines, as shown in the diagram of 
solid and dotted lines, as Q7° and Q' 7°°. 



On the horizontal lines and measuring from Q, 
place the distance of the dotted lines in elevation, like 
11 13 in elevation is Q 13° in the diagram of dotted 
lines, and so on. Similarly, 16 9, for in- 
stance, is Q" 16 00 in the diagram of solid 
lines. Carefully connect proper points, in 
both diagrams, on the horizontal line with 
those on the vertical lines so that the lines 
will always have the same number designat- 
ing the ends, though they have dissimilar 
prunes. 

Fig. 763 is the full net pattern of the 
boss and the first step in its development is 
to locate at pleasure the point 19". Using 
this point as a center describe short arcs, 
setting the compasses for same successively 
to the distances prescribed in the diagram 
of lines of the diamond part. To illustrate 
this point, 19" 1" in the pattern, Fig. 763, 
equals 19 s I s in drawing, Fig. 762. Starting 
on the arc at point l xx step to the next arc 
by the use of the dividers, the space 1' 2' in 
diagram C; then from this point so obtained, 
as 2 XX in the pattern step to the next arc, 
the distance 2' 3' of diagram C, and so forth. 
Inasmuch as a full pattern is required, this 
process is to be repeated for the other half 
of the diamond, each move of the process 
simultaneously for each half. 

After having obtained line 5 XX 19 xx in the 
pattern, the curved part of the boss is de- 
veloped by the customary procedure of 
triangulation with clotted and solid lines; 
that is to say, from 5 XX describe a short arc 
toward 19" that has a length coincident 
with 7° 18° of the dotted line diagram. In- 
tersect this arc with one swung from 19" that 
has a length equal to the space 19 18 of the 
circle in the elevation. 

From 18" and toward 5 XX swing an arc 
of a length coincident with the line 18°° 7°° 
of the solid line diagram. Intersect this arc 
with one from 5" of a length equal to the 
space 6' 7 s of diagram D. Repeat these 
steps until line 12" 13" is obtained. Then a 
line traced through the points of intersection 
gives the pattern for the boss. It must be remembered 
that like spaces in diagram D are carried successively 
to the pattern and the circle spaces, like 19" 18" and 
so on, are always the same. 



Pattern Problems. 



437 



In forming up the boss, an acute bend is made along 
line l zx 19". Also slight bends are made on lines 5 ZZ 19" 
the cross hatch diagrams M M of the pattern indicating 
the direction in which the bends are to be made. The 
bends on lines 5 TI 19 zx are to lie approximate and prac- 



boss, while bend on line l xx 19" is made to the angle 
ABC. This angle is found by drawing a horizontal 
line anywhere equal in length to line 5 19 6. At the cen- 
ter erect a vertical line equal in length to line 5 X b. 
Draw lines to points so acquired as A B C, which gives 



tically rubbed out in rounding up the curved part of the | the desired angle. 

PROBLEM 226. 
Pattern fcr Round to Rectangular Elbow. 



A correspondent who experienced some difficulty in 
solving a pattern problem recently sent in a request 
for assistance. His letter and the solution of the prob- 
lem are given in the following: 

L M N 

K 
Elevation 

J, 



m' q 





Fig. 76Jf. — Elevation and Profile of an Elbow 
from Round to Rectangular as Illustrated 
in Problem 213. 



Both myself and a friend are fond possessors of 
"The New Metal Worker Pattern Book," but even 
with every rudiment of triangulation before us we are 
stuck on one pattern problem which is the only one in 
the book .that has caused us 
trouble. We are therefore seeking 
a little assistance in solving 
"Problem 213," which consists 
of connecting an elbow with a 
round and a rectangular pipe. 
The elevation is very simple to 
obtain from the explanation. To 
make myself clearly understood, I 
mean that portion of the eleva- 
tion which is G G' and F F' as 
shown in Fig. 714 in the book, 
but when it comes to the handling 
the section L L' and K K', there 
is a difference, and although we have studied and 
worked on this part of the problem persistently we can- 
not seem to understand how to work out the pattern. 
We would therefore be very grateful for some more 
detailed information as to how to proceed than we get 
from the book. Having mastered all the rest of the 
problems in the book, we do not wish to be stuck by 
this one, and hope to get the additional explanation 
which we seem to need. 

Answer. — With this problem, as with all others 
which must be solved by the triangular method, abso- 
lute concentration is essential to obtain the correct result. 
If one step is misapplied the balance of the work is 
in error and a new start must be made, for the mis- 
take is difficult of detection. A careful perusal of the 
text of "Problem 213 " referred to shows that the procedure 
there laid down is not at fault, but possibly too much 
of the working out of the problem has been left to the 
reader. 



438 



The New Metal Worker Pattern Book. 



There are seven sections to the elbow of this prob- 
lem, and every one has a different pattern. The pat- 
terns for the first and last sections, as stated in the 
book, are simple to obtain, and information is also pre- 




Fig. 765. — Elevation of One Section Showing Methods of Triangulation 
as Illustrated in Problem '213. 



sented for obtaining the pattern for the section which 
joins the round pipe. If the correspondent thoroughly 
iinderstands the procedure up to acquiring this pattern, 
the next operation is to obtain true profiles of the third 




Fig. 766. — Elevation of Third Section. 

section. There is presented herein a new illustration, 
Fig. 766, which gives the elevation of the third section. 
The bottom profile of this section must be an exact 
duplicate of the top profile of the second section in 



Fig. 765, inasmuch as the profiles must correspond to 
have sections join each other correctly. Fig. 765 is a re- 
production of Fig. 718 of Problem 213. 

Following the text, Problem 213, for obtaining the 
profiles, the top profile of Fig. 766 is developed. The curve 





A' . 


/ 


/ 








— 


h 


4 


z. 














z' 


J: 


\o 


p 


J 


J' 


f 


V 


,' 




■\\ 






,..■■ 






if 



Fig. 767. — Elevation of Fourth Section. 

X to R is drawn at pleasure — the points X, R, R', and 
X' being fixed, the intermediate lines are arbitrary. The 
pattern of this section, Fig. 766, is obtained as exem- 
plified in Problem 213 by Fig. 718, and the procedure 




2 A 
Fig. 768. — Elevation of Fifth Section. 

repeated for the fourth section, as demonstrated in Fig. 
767; and again for the fifth section, as in Fig. 768. 

In Fig. 769 is shown the sixth section, which gave 
trouble to the correspondent. Having drawn the ele- 
vation as L, L', K' and K with the bottom profile dupli- 
cated from the top one of Fig. 768, and dividing the 
round portions into three equal spaces, giving the points 



Pattern Problems. 



439 



K, B, 5, 6, D, D', 7, 8, B', and K', the next step is to 
draw vertical lines from these points of division to the 
line K K', and labeling the points of intersection respec- 
tively r, s, k, k', t, and u. On the top of the section 




Fig. 769. — Elevation of Sixth Section. 

from points L and L', and at right angles to this line, 
lines are drawn equal to the distance N n of the profile 
of the rectangular pipe m Fig. 764, and the ends of the 
lines marked M and M' and these points connected 
with a line which gives the true profile on the hne L L' 
of the section. As will be seen, the transition is now 



drawn, equal in length to the hne k' D' of Fig. 769 and 
designated k' D'. A vertical line is drawn, equal in 
length to k L' of Fig. 769, and so marked. Parallel 
to k' D' and from L' in Fig. 770, a line is drawn, equal 
in length to L' M' of Fig. 769. Connecting points M' 
and D' of Fig. 770 result in a line the correct length of 
k' L' of Fig. 769. Tins is repeated for all the lines in 
Fig. 769, as shown by hke numbers and letters of Fig. 
770. 

For the pattern proceed in this fashion: Anywhere 
redraw P 1 of Fig. 770, as shown by L M K B in Fig. 
771. Using M as center, swing arcs successively equal 
in length to M 5, M 6, and M D of Fig. 770, as shown 
in Fig. 771 by the same symbols. From B of Fig. 771 
step to each arc with the compasses set to the space 
B 5 of Fig. 769. Again, from M of Fig. 771 as center 
and compasses set to the distance M M' of Fig. 769, 
describe an arc. Using D now, of Fig. 771, as center 
and compasses adjusted to the distance D M' of A P 
of Fig. 770, intersect the arc just drawn, obtaining in 
Fig. 771 the point M'. Still using D of Fig. 771 as 
center and compasses set to the space D D' of Fig. 
769, describe an arc, which intersect by one drawn of 
a length equal to M' D' of A 1 of Fig. 770 and using 
M' as center. With M' as center draw arcs of a length 
decreed by A 2, A 3, and A 4 of Fig, 770, which are 




Br Si 6 K D 

Fig. 770. — Method of Getting True Lengths of Various Lines of Sixth Section. 



from a rectangle with approximately quarter-circle cor- 
ners to a rectangle, perforcing a change from the process 
given in Problem 213 for the other section in the de- 
velopment of the pattern of the section. 

From r, s, and k lines are converged to point L; 
also from k', t, and u to L', all as indicated in Fig. 769. 
To assist further in developing the pattern, a dotted hne 
is drawn from k to L'. 

To avoid deviation from the style of Problem 213, 
the correct lengths of the various lines in Fig. 769, as, 
for instance, line k' L', is ascertained by the method 
indicated in Fig. 770, as follows: A horizontal line is 



stepped, as before, by the dividers set to the space 
D' 7 of Fig. 725. The rest of the pattern is A 4 of Fig. 




B 5 6 D 



Fig. 771. 



726 duplicated, but reversed; the transferring of this 
part of the pattern can be accomplished by tracing or, 



440 



The New Metal Worker Pattern Book. 



better still, triangulation, as was followed for the part 
of the pattern M M' D' and D. 

When forming up the pattern a slight crease or 
bend must be made on the line D M' of Fig. 727, inas- 
much as D, D', M', and M do not lie in a flat plane. 



This apphes to the flat parts of the sides of the third, 
fourth, and fifth sections, as shown by the dotted lines 
in Figs. 724, 723, and 722. For some reason or other 
this is not stated in the book, possibly because the bend 
is so slight as to be imperceptible. 



PROBLEM 227. 



Four-piece Tapering Elbow. 



In the problems relating to two- three- and five- 
piece tapering elbows which are given in the preceding 
pages, it is assumed that the various pieces are to have a 
regular taper, so that when the pieces are revolved in 
a certain manner they will constitute the frustum of a 
regular cone. It is then only necessary to develop the 
patterns in the customary manner for radial line problems. 

The following description of a procedure for a four- 
piece tapering elbow is along the same fines. In all elbow 
problems the first thing is to find the rise of the miter line. 
There are several ways of doing this, and about as good 
a way as any is given in Fig. 772. The procedure is to 
describe a quadrant of any convenient radius, as A G H, 
H being the center. This quadrant is divided into six 
equal spaces, as A to G. Drawing lines through the 
points B D F shows the four pieces of the elbow, so that 
A to B is the rise of the miter line B H. 

Assume in Fig. 773 that the half circle 1 to 7, with 
Z as its center, is the half plan of the large end of the 
tapering elbow, which has been divided into six equal 
spaces; the more spaces used, of course, the more accurate 
the final results. These points in plan are projected to 
the base fine, l x T of the elevation, continuing the center 
line indefinitely. This base fine is extended somewhat 
to the right and a point H, located where the judgment 
tells that throat T L is of sufficient size ; that is, the farther 
away the point H is from point T the more throat there 
will be; while if point H is nearer point T there will be 
less throat or space 7 X to L. 

Having located point H, the compass is set to radius 
H A of Fig. 772, and then with H, in Fig. 773, as center, 
an arc is struck. Then the compass is set to A to B of 
Fig. 772 and with A of Fig. 773 as center, an arc is described 
cutting the one previously described, giving point B. A 
line is drawn from H through B and well to the left, 
crossing the center line at 4'. Point 4° is the same dis- 
tance from 4' as 4' is from 4 T . 

Through point 4° draw a horizontal fine extending 



it to the left. Locate point H' the same distance from 
4° that H is from 4 s . Locate point A', then B', as A 
and B were located. Draw a line from H' through B' 




*f-=~tf Fig. 773. 



Fig. 772. — Finding Rise of Miter Line. Fig. 773. — Diagram and 
Patterns of a Regular Tapering Four-Piece Elbow. Fig. 774- — 
Elevation of Four-piece Tapering Elbow with Pieces in Correct 
Position. 

as shown. Repeat these operations, and thus obtain the 
other miter line B" H". Also locate point 4°° the same 
distance from 4 XI that 4' is from 4 X , and draw a horizontal 
line through 4°°, making this line— l xx T x — the width of 
the small end of the tapering elbow. Draw a line through 
point l x from l x until it touches the center line at Z x . 



Pattern Problems. 



441 



Do the same from T through 7" to Z T , which completes 
the cone, with the various miter lines, and so forth, 
inscribed on it. 

Using Z 1 as the center and Z x T as radius, describe 
a long arc, as 1°. Step off on this arc the space 1 to 7 
of the plan, as 1° to 7°. Lines are drawn to these points 
from the apex of the cone, Z x . From where the points 
1 to 7 in plan are projected to the base line l x T element 
lines are drawn, as shown by the dotted lines, to the apex 
Z X . Where these element lines cross the miter lines 
mark intersection points. Then, from these intersection 
points on the miter lines horizontal lines are projected 
across the line T x T as shown. Now, from these points 
lines are swung around, using point Z 1 as center, to like 
element lines for the pattern. For instance, point 4' 
is projected horizontally across to 4 B thence swung to 
line 4° in the pattern, giving point 4 BB , and so on. The 
arc from 7 XX , of course, is the end line from the top piece, 
and lines traced through the other points — like 4 BB — 
complete the one-half pattern of all the. pieces. 

It is to be noted that in this procedure there is no 
way of predetermining the positive dimensions of the 
elbow over all. It can be told what the throat, like T L, 
will be, and also what the small and large ends will be 
as required, but that is all. In Fig. 774 is portrayed the 
appearance of the elbow when the pieces are joined 
together in correct relationship to each other. This 
drawing was made by tracing each piece separately in 
Fig. 772 and transferring them, as they belonged, in 
Fig. 774. 

In all such problems having the end pieces tapering 
it is highly probable that some sort of a straight collar 
must be joined to the end pieces, as shown in Fig. 774. 
Obviously, the object then becomes a six-piece elbow 
instead of four-piece. With these foregoing facts in 
mind it is customary nowadays to make the end pieces 
in all such elbows straight, and have whatever taper, 
offset, or other such requirements, provided for in the 
intermediate pieces. 

A four-piece tapering elbow, designed with these 
ideas in mind, is shown in Fig. 775. Knowing the dimen- 
sions of the large and small ends, the general dimensions 
over all can be determined on and followed out in the 
designing of the elbow. So then, as for the other pro- 
cedure, the first thing to do is to find the rise of the miter 
line. Let it be supposed that the base line of the large 
end is 7' 1' of Fig. 775, and that the line of the small end 
is 7" 1", situated at right angles to the base line, and 
that it is required that they be that much away from 
each other. 



These two lines are extended to meet as at H. From 
H, as center, a quarter circle is described from 1' to 1" 
and this arc is divided into six equal spaces. Lines are 
drawn from H through the first point of division on the 
arc, the third and the fifth division, giving thereby the 
miter lines of the various pieces. 

A half profile of the large end is drawn, also one 
for the small end as shown. These half profiles are 
divided into equal spaces, say six, and fines are projected 
from the division points to the miter lines of the end 
pieces, incidentally completing the outlines of the end 
pieces. Just what dimensions the two intermediate pieces 
should have is a matter of choice. In this design the 
throat lines are already determined by the arc, as shown; 
now, a good method for deciding on the length of the 
miter line I s 7 B is to place the distance l c 7 C on this line, 
giving point Z. Also place distance 1 E 7 E on this line, 
giving point X. Divide space Z X into halves, and' point 
7 B will be the extreme end of that miter line. Connect 
points 7 C 7 B and 7 B as shown by lines h and b, completing 
the elevation of the elbow. 

It is not necessary to develop the profile on lines 
7 E 1 B and 7 C l c , but it must, however, be developed on 
line 7 s I s . Point A J is midway from l B and 7 B . Draw 
a line from i J at right angles, as shown. Here again 
is a certain dimension a matter of choice. The widths 
across at points 4 C and 4 B are fixed, but at 4 J can be at 
will. So, take the space 4 Y of the profile of large end 
and place it as 1 J Y'. Also space in profile of small end 
of 4' 4 Y x , as 4 J Y°. Then 4 B is midway between Y° and 
Y'. Sketch the profile 7 B i B 1 B in free hand and divide 
into six spaces as shown. Draw fines parallel to 4 J 4 B 
from these points as shown, to the miter line. Connect 
the various points on all miter lines with a series of solid 
and dotted lines, in the customary manner, as the letters 
from A to W show. 

The half pattern of the large end is shown to the 
right of Fig. 775, and is an ordinary parallel line develop- 
ment, the stretchout being taken from the half profile 
as the large end and points l c to 7 C projected to this 
stretchout. This same procedure is followed for the half 
pattern of the small end shown below point H. 

The next thing to do is to find the true lengths of 
lines A to W; so, in Fig. 776, draw a horizontal line, 
and measuring always from point 1-7 in Fig. 776, place 
the distances or lengths of lines in Fig. 775 for piece 
No. 2 as U S Q M'. Erect vertical lines as shown. 
On line from point 1-7 in Fig. 776 set off the lengths 
4 .4 y* 3^ 3^ anc i 2 A 2 P from small profile in Fig. 775. 
Then, taking the distance A J 4 B in Fig. 775, place it on 



442 



The New Metal Worker Pattern Book. 



line from point Q' in Fig. 776. Space S J 3 B of Fig. 775 on 
lines from points S' and 0' of Fig. 776, and space 2 J 2 s 
of Fig. 775 on lines from points 0' and M' of Fig. 776, thus 
obtaining points 2 B 2, B 4 B 5 B and 6 B in Fig. 775. Con- 



L, point 1-7 goes to 2 B of Fig. 777; point 6-2 to 3 s and L; 
point 3-5 to 4 s and 6 B , and point 4 to 5 s . Exactly the 
same process applies to finding the true lengths of lines 
A to K of piece No. 3 of Fig. 775, as given in the diagrams, 




Fig. 781. 



Fig. 775. — Diagrams and Patterns of Four-piece Tapering Elbow with Parallel Ends. Fig. 776. — Diagram of True Length of Solid Lines 
for Piece No. 2. Fig. 777. — Diagram of True Length of Dotted Lines for Piece No. 2. Fig. 77S. — Diagram of True Length of Solid 
Lines for Piece No. 3. Fig. 779. — Diagram of True Lengths of Dotted Lines for Piece No. 3. Fig. 780. — Method of Obtaining 
Pattern for Piece No. 2. Fig. 781. — Method of Obtaining Pattern for Piece No. 3. 



necting these points to like numbered points in Fig. 776 
gives the true lengths of lines USQO and M. 

The procedure, as given in Fig. 777, is about the same 
for the dotted lines, except, however, instead of con- 
necting like number points to get lines WTRPN and 



Figs. 778 and 779. Note that the spaces on line from 
point 1 B -7 B of Fig. 779 are taken from the profile in the 
elevation of Fig. 775, like l B to 2 B of Fig. 779 equals 
space 2 J 2 B of Fig. 775, and spaces from points J' to B' 
of Fig. 779 are taken from the half profile of the large 



Pattern Problems. 



443 



end, as space F' 4? of Fig. 779 equals space 4 Y of Fig. 
775. Space H' 2 J of Fig. 779 equals space 3 3 r of Fig. 
775, and so on; and just so in Fig. 780 for the dotted 
lines, except, as for Fig. 777, different numbers are con- 
nected as shown. 

The pattern for piece No 2 is obtained as in Fig. 780. 
Draw the vertical line b, in Fig. 780, equal in length to 
line b of Fig. 775. Set the compass to the true length 
of line L in Fig. 777 and describe a short arc, in Fig. 780, 
using point 7 as center. From point 7 A in Fig. 780 
describe an arc to intersect the arc just drawn, this arc 
to be of a radius equal to space 7°°-G ° in the pattern of 
the small end in Fig. 775. Using 6* 1 in Fig. 780 as center, 
describe an arc equal to the true length of line M as 
given in Fig. 776. From 7 in Fig. 780, as center, describe 
an arc to intersect the one from 6 A , of a radius equal 



to space 7 B Q B of the half profile, 7 B to I s as given dotted 
in Fig. 775. Continue in this manner until line a of 
Fig. 7S0 is reached, which is the same length as line a 
in Fig. 775. This process completes the half pattern for 
No. 2 piece. 

The pattern of piece No. 3 is given in Fig. 781. 
Starting from the vertical line h, which equals line h 
of Fig. 775, the true lengths of lines are then taken alter- 
nately from Figs. 778 and 779. The spaces, as 7 to 6, 
are taken from the clotted half profile of Fig. 775, as 
7 s to Q B , and so on, and the spaces 7° to 7°, for instance, 
are taken from the pattern of piece No. 4, as 7°° to 6°°, 
and so on to line r of the pattern, which equals line r 
of Fig. 775. If, while developing the patterns from line 
h toward line r, the process was repeated in unison to the 
other side of line r, a full pattern would be acquired. 



PROBLEM 228. 



Developing a Double Offset Boot. 



By this procedure the front and side elevations are 
not necessary in the development of the pattern, but 
are here shown to faciliate the illustration. In all pro- 
jection drawings of this kind the center line only is 
dealt with, projecting this line to the various views, 
and then build the article around this center line, so 
to speak. 

The first step is to draw the angle and the center 
line of the boot in the front elevation as shown by A B C, 
making the length of the center line of the rectangular 
pipe as desired as shown by A B, the vertical height 
of the center hne of the transition piece as B a and its 
projection as much as a C, thus giving the foreshortened 
view B C. 

Extend A B in the front elevation, and upon it 
establish the point B, which represents the plan view of 
the vertical hne A B in elevation. At right angles to 
A B draw the line B /, which intersect by a line dropped 
vertically from the point C in the front elevation, ex- 
tending this line as shown by / C, or the amount desired 
which C in the front elevation shall lean toward the 
reader from the vertical hne A B, as indicated from 
/ to C in plan. The line C B in plan then gives the 
foreshortened plan view of the center hne of the tran- 
sition piece. 

As this center line of the boot does not show its 
true angle or true length in either plan or elevation, 



a true elevation must be constructed which will show 
the true angle and length of the center line, as follows: 
Parallel and equal to B C in plan, draw any line as a C 
in the true elevation and from a erect the perpendicular 
a A, upon which place the distances a B, and B A, 
equal, respectively, to the distances a B and B A in the 
front elevation. Draw a hne from B to C in the true 
elevation, then will ABC show the true lengths and 
true angle of the center line of the boot. 

At right angles to B C in the true elevation, through 
the point C, draw the line 6 10 equal to the diameter 
of the round end of the transition piece. Obtain the 
true miter line of the boot, by using B as center and 
with any radius, describe the arc, cutting the hnes A B 
and B C at c and d respectively, which use as centers and 
with the same or any other radius, intersect arcs at e. 
Draw the miter line through e and B indefinitely as 
shown. 

Construct around the center point B in plan the 
profile of the rectangular pipe as shown by 2, 3, 4, and 
5, being careful that B will be directly in the center 
of the rectangle. Bisect the side 2 5, and obtain 1, 
which represents the seam or joint of the rectangular 
pipe. From the various points 1, 2, 3, 4, and 5, parallel 
to the line A B in the true elevation, project hnes until 
they cut the miter line e B extended at 1, 2, 3, 4, and 
5. Through A draw the line H G at right angles to 



444 



The Neiv Metal Worker Pattern Book. 



Front Elevation . 



Side Elevation. 



True Elevation . 




Fig. 783. — Geometrical Method of Projecting and Developing Double Offset Bool. 



Pattern Problems. 



445 



A B and complete the true elevation of the rectangular 
pipe 5 G H 3. 

Extend the center line B C in the true elevation, 
and upon it establish any point as D, which use as a 
center and describe the circle shown, equal to the diam- 
eter 6 10, and divide the circle into, say, eight equal 
spaces, as shown from 6 to 13. In practice more spaces 
should be employed. Through these various figures in 
D draw lines parallel to D C, cutting the plane 6 10 in 
the true elevation as shown by similar numbers. Draw 
lines from 6 to 3, and from 5 to 10. Then H, 3, 6, 10, 
5, G shows the true elevation of the double offset boot. 
If the line 6 10 in the true elevation were drawn at a 
different angle to the center line B C, or in case the 
line 6 10 were drawn parallel to the miter line e 5, thus 
placing the round pipe parallel to the rectangular pipe 
H B 5 e and making a double elbow in the offset, then 
the principles which will follow in drawing the plan 
and projecting the various views would be similar. The 
true elevation just drawn forms the basis for projecting 
the plan, front, and side elevations. It makes no dif- 
ference what profile the pipe may have, whether round, 
square, elliptical, or rectangular, or at what angle the 
boot is drawn in either plan or elevation, the principles 
which will follow are applicable to either case. 

The plan view is completed by extending the center 
line B C and upon it establishing the point D°, which 
is used as a center to describe the circle shown, of sim- 
ilar size to D in the true elevation. At right angles 
to B D° in plan, through the center D° draw the diam- 
eter as shown. Divide the circle D° into the same 
number of spaces as D, being careful to give the circle 
D° a one-quarter turn so as to bring the point 6 and 
10 on the center fine B D°, while in the profile D the 
points 6 and 10 are at right angles to the center line. 
The reason for this quarter turn in the circle D° be- 
comes evident when referring to the true elevation, in 
which the points 6 and 10 are at the top and bottom 
of the transition piece and come directly on the center 
fine of the boot when viewed in the direction of the 
arrow. 

Number the points in the circle D° as shown from 
6 to 13, and from these points parallel to D° B draw 
lines, which intersect by lines drawn from similar num- 
bered intersections on the plan 6 10, in the side ele- 
vation, at right angles to B D° in plan. Through the 
points of intersection 6 to 13 in plan the elliptical figure 
is traced, having for its center the previously estab- 
lished point C. 

Connect the flat surface of the rectangular pipe 2 3 



to point 13; 3 4 to 7; 4 5 to 9; and 5 2 to 11, and as- 
sume that the seam of the transition piece will come 
along 1 11. From point 6 draw a clotted line to 3; 
from 8 to 4; 10 to 5, and 12 to 2. This completes all 
the drawings that are necessary in the development of 
the patterns. 

Parallel to G H in the true elevation draw any fine 
as F E. Upon this place the length of the fines in plan 
as shown by similar numbers on F E. For example: 
Take the distance of the seam line 1 11 in plan and 
place it as shown by 1' 11' on F E. At right angles 
to F E from points 1' and 11' draw lines, which inter- 
sect by lines drawn parallel to H G in the true eleva- 
tion from the point 1 on the miter fine 3 5 and from 
the point 11 on the plane 6 10, thus obtaining the inter- 
sections 1 and 11 in the true lengths. Draw a line 
from 1 to 11 in the true lengths, which gives the true 
length of the line 111 in the plan. 

In a similar manner take lengths of 2 11, 2 12, 
and 2 13 in the plan, and place them on the line F E 
in the true lengths as shown by 2' 11', 2' 12', and 2' 13'. 
From these points perpendiculars are drawn and inter- 
sected by lines drawn from similar numbers in the true 
elevation, parallel to F E or H G. In tins manner 
all of the true lengths shown by the solid fines are 
found. 

Before the transition pattern can be developed it 
is necessary to develop the pattern for the rectangular 
pipe. From the miter cut the true edge line of the 
upper end of the transition piece. Take the girth of 
the rectangle from 1 to 5 to 1 in plan, and place it as 
shown, from 1' to 5' to 1" on the vertical fine H° G°. 
At right angles to H° G° through the small figures 
draw fines indefinitely as shown. Now, measuring from 
the fine H G in the true elevation, take the various 
distances to points 1, 2, 3, 4, and 5 on the miter fine 
e 5, and place them on fines drawn from similar num- 
bers, measuring from the fine H° G°, thus obtaining ' 
the intersections 1°, 2°, 3°, 4°, 5°, 1°. Lines connect- 
ing the points gives 1' 1° 1° 1", the pattern for the 
rectangular pipe. 

The pattern for the transition piece is obtained as 
follows: Take true length 1 11, or the seam length, 
and place it as shown by 1° 11 in the pattern. With 
1° 2° in the rectangular pipe pattern as radius, and 
1° in the transition pattern as center, draw the arc 
2°, which intersect by an arc, struck from 11 as center, 
and with 11 2 in the true length as radius. Now, with 
radii equal to 2 12 and 2 13 in the true lengths, and 2° 
in the transition pattern as center, draw the arcs 12 



446 



The New Metal Worker Pattern Book. 



and 13. Now set the dividers equal to the divisions 
in the circle D or D° and, starting from 11 in the tran- 
sition pattern, step to arc 12, and then to 13. 

Continue in this manner until the entire pattern is 
developed. No mistake can occur if care is taken; the 
distances from 1° to 5° to 1° in the transition pattern 
are obtained from 1° to 5° to 1° in the rectangular pipe 
pattern; the distances from 11 to 11 in the transition 
pattern are obtained from the spaces in the profiles D 
or D°, while the straight lines in the transition pattern 
are obtained from the true lengths. Trace a line through 
points thus obtained in the transition pattern, then will 
1° 11 11" 1° be the desired pattern. To this edge must 
be allowed for seaming. This, then, completes the pat- 
terns for a boot taken off the corner at an incline, the 
area of which is maintained. 

Take a tracing of the true elevation, with the vari- 
ous points of intersections on it, and place it to the 
left of the center line A B in the front elevation as 
shown, being careful to have G 5 run parallel to A B. 
From the various intersections 2 to 5 in the true ele- 
vation, draw horizontal lines to the right indefinitely. 
Intersect these by vertical lines erected from points 
2 to 5 in the plan, thus obtaining the intersections 2 
to 5 in the front elevation. In a similar manner, from 



points 6 to 13 in the true elevation, draw horizontal 
lines to the right indefinitely and intersect them by 
vertical lines, erected from similar numbers 6 to 13 
in plan, thereby obtaining the intersections 6 to 13 in 
the front elevation. Trace fines through points thus 
obtained and connect the flat surfaces of the rectangular 
pipe, similar to that shown in the plan, which com- 
pletes the front elevation. Notice that the transition 
piece has been drawn around the center line ABC 
completing thus, the full projection drawings of the 
object. 

For a side elevation, trace the plan in the position 
shown by X, with the various points of intersections 
marked. From these vertical lines are erected, inter- 
secting those of similar numbers previously drawn from 
the true elevation. A line traced through points so 
obtained will give the side elevation, all clearly shown 
by similar reference numbers. So simple is this method 
in projection that if a view or elevation were desired 
looking directly into the center of the circular opening, 
as C in the plan, it would only be necessary to turn the 
plan so that the fine C B would run vertically, and 
then project the various points in the plan to meet the 
similar numbered lines drawn horizontally in the ele- 
vation. 



PROBLEM 229. 
Pattern for Straight Back Transition Piece. 



The following is an explanation of the method of 
developing the pattern for an object changing in shape 
from square to round, and having a straight back. 

This change is made from a rectangular to a cir- 
cular shape within a certain distance or bight and in 
one line on one side. The pattern for such an object 
is developed by the branch of surface of solids devel- 
opment technically known as triangulation and involves 
the elementary technique of that science. 

On page 327 a similar problem is presented, except 
that the round shape is situated centrally in respect to 
the cross axis or center lines of the square, or rather 
rectangular shape, while here the one side of the square 
is tangent to the circle and the center of the circle is 
on the long axis of the rectangle. 

In that problem the object is what is termed an 
object of symmetrical quarters, while here it is an object 



of symmetrical halves. The procedure for cutting the 
pattern is identical in both cases, only with quarters alike, 
the pattern for one quarter will do for the whole, whereas, 
in this problem, the pattern must be acquired for one 
half to do for the whole of the object. 

This does not change the procedure described on 
page 327, but requires more labor, because two sets of 
triangles, instead of one, must be laid out. By laying- 
out is meant that the plan of the object gives the base 
lines of a certain set of triangles and the elevation 
gives the altitude and the hjqjotenuse or true length 
of line, which is what is wanted, is found by constructing 
these series of triangles and plotting out, as it were, 
these true lengths of lines in correct relation to each 
other, which develops the pattern. 

The plan view of the object is shown in Fig. 783. 
One half of the circle is divided into equal spaces and 



Pattern Problems. 



447 



lines drawn from these points, as to 8, to the corners 
A and B. These lines are the base lines of the triangles 
mentioned and an elevation is projected from the plan 
as shown, which gives the altitude C D of the triangles. 

Elsewhere, as in Fig. 784, draw a line indefinitely 
to represent the base lines of the triangles. At right 
angles to this line, inasmuch as these generally are right 
angle triangles, draw a vertical line to the length of 
C D of Fig. 783, which is the altitude of the set or series 
of triangles. From A of Fig. 7S4 set off the lengths 
A 0, A 1, A 2, A. 3, and A 4 of Fig. 783. Then con- 
nect to D as shown. Do the same hi Fig. 785 for the 
base lines B 4, B 5, etc., of Fig. 783. 

To develop the pattern draw a line, anywhere it is 
convenient, of a length equal to C D of Fig. 783, as 
C in Fig. 786. Set the compass to the space A 
in Fig. 783, and with the point of the compass at C in 
Fig. 786, and swinging to the right describe a short 
arc. With the compass set to the space O D in Fig. 
784 and its point at in Fig. 786, describe an arc inter- 
secting the first arc, establishing point A. Now, with 
the point of the compass always at A, describe arcs 
of a radius of 1 D, 2 D, and so on of Fig. 784; then, 
with the dividers set to the division space of the circle 
to 1, etc., of Fig. 782, step from point in Fig. 786 
to arc 1, then to arc 784, then 3 and 4 and from the 
points so located draw light lines to A. 

From A in Fig. 786, describe an arc to the right of 
a radius of A B in Fig. 783. From 4 intersect this arc 
with one of a radius of 4 D of Fig. 785, giving point 
B. Connect the points as shown and then with the 
point cf the compass always at B describe arcs of a 
radius of the lines 5 D, etc., of Fig. 785. Step these 
with the dividers as before and draw the connecting 



lines; also sketch a line through points to 8. From 
8 swing an arc of a radius of E F of Fig. 783, and inter- 
sect with an arc of a radius of B G of Fig, 783, estab- 




Fig. 7S6. 



Fig. 783. — Plan and Elevation of Object. Figs. 784 and 785. — True 
Length of Lines in Plan. Fig. 786. — Half Pattern for Object. 

lishing point G. Connecting B to G and 8 to G with 
lines completes the pattern. Note that angles C and 
G must be right angles; if not, the pattern is incorrect. 



448 



The New Metal Worker Pattern Book. 



SECTION 4. 

m®d ©f C©mMiraa{di©ini Formal 



Tn the other three sections of this chapter, extreme 
care was used to properly enter the problems in the 
right section according to their class. This was done 
principally to assist students in then* progressive study 
of the science. Classification, however, is of little mo- 
ment when the book is used for reference, as for instance, 
a pattern cutter who has an offset boot, like Problem 209, 
to lay out, would not need to be told which of the three 
methods of surface development should be employed. 

Another consideration is the fact that most prob- 
lems may be solved by more than one method. In 
the problem mentioned, for instance, not only is tri- 



angulation resorted to, for the development of the 
middle piece, but the parallel line system would also 
be used to develop the patterns of the collars. So then, 
rather than attempt to classify the many modern prob- 
lems that are to be added to the revised edition and 
perhaps confuse the well-tried arrangement of the former 
edition, these new problems are all grouped in this Fourth 
Section containing what is confidently felt to be the most 
valuable collection of modern problems that it is pos- 
sible to gather. Always the leading book for reference 
or home study, this new edition gives the most complete, 
finest and largest compilation of problems obtainable. 



PROBLEM 230. 



Patterns for Bosses for Sheet Metal Hand Pumps. 



In Fig. 787 is given the illustration of a hand pump 
with reinforcing bosses and an explanation of the method 
of developing the patterns of the bosses designated by 
the letters a and b is given herewith. A study of the 
problem shows that there is a cone, which is the funnel 




Fig. 787 — The Problem. 

part of the pump and a cylinder, the spout, which inter- 
sects this cone. The axis of the cylinder is at right 
angles to the axis of cone, and to reinforce the connection 
of the cylinder with the cone, two bosses are joined to 
them. 

The information gathered anent this object shows 
that each maker has his own idea as to proportion of 
parts, but all makers, it seems, use the double boss, 
like the illustration. Why they do not employ a single 



conical boss as is customary for reinforcing the faucet 
to a can or other such objects, is a mystery; for the 
double boss 011I3' gives vertical stiffness, whereas the 
conical boss would give both lateral and vertical rigidity; 
that is to say, lend resisting ability to the spout to side- 
wise pull from the funnel, as well as up and down. 

That, however, is beside the question, and there- 
fore, in Fig. 788 is shown a general view of the object 
proportioned as nearly as possible for so small an illus- 
tration from available data on the subject. The process 
of determining the miter fine, or rather, intersection fine 
of the cylinder with the cone, has been exemplified 
herein, so that it will be passed over, except to say 
that Fig. 788 should be sufficiently clear to be self- 
explanatory in respect to that procedure. As was 
stated previously, each person has his own idea of de- 
signs for this object, so it is a matter of choice as to 
how the bosses should be designed. The bosses in 
Fig. 788 give the idea of one party as to what is about 
right, and should they be wanted smaller or larger, 
or so on, the methods hereafter expounded and recom- 
mended for this case can be applied with but slight 
adjustment, to whatever variance there may be. 

However, point A in Fig. 788 is essential, so before 
proceeding, note that point 3 in elevation profile of 
the spout is projected to the side of the cone at B then 



Pattern Problems. 



449 



down to line 6 T in the plan, as at B'. From T swing 
an arc from B' and where it crosses line 9 locate thereby 
point A'. Project A' up to line 9 3 of elevation, getting 
thereby the required point A. 

As there are two bosses, two separate operations 
are requisite for developing the necessary views and 
then the patterns; also, as many work lines and the 
like are utilized, individual illustrations were prepared. 
And again, Fig. 789 shows how the pattern was 
acquired for boss b of Fig. 787. The cone is a repro- 
duction of the one in Fig. 7S8, also the cylinder. Point 
A of Fig. 788 is carefully located in Fig. 789 by trans- 



same element line in plan, as at R'. Proceed that way, 
as shown, acquiring thereby the section lines in plan 
R' and S'. 

Place the profile of the cylinder in plan as shown. 
Project point 2° to section line S', thence up to line 
2' 2", giving miter line point 2 K. Project point 1° 
in plan profile to section line S', thence up to line 1' 1" 
getting point 1 M. Trace a line through points o", 
1 M, 2 K, and A, which is then the sought-for miter 
line, or, rather, line of intersection of cylinder with cone. 

A true section of the boss must be developed as fol- 
lows: Draw line X Y at right angles to lines from the 




Fig. 788. — Plan and Elevation of Spoul 
and Re-enforcing Bosses. 



Fig. 789.— Pattern for Boss b 
in Fig. 7S7. 



Fig. 790.— Pattern for Boss a, 
Fig. 7S7. 



ferring, say, distance A B of Fig, 788 to A B of Fig, 
789. At will, place point o' on the cylinder, also point 
o" on the cone. Draw straight lines from o' and o" 
to A. Divide profile of cylinder into, say, six equal 
spaces. Project lines through points o to 3 to line o' A 
and thence indefinitely through the cone parallel to 
fine o' o" , as shown. 

Divide the quarter plan of cone into a number of 
equal spaces and draw fines from those points to T, 
which gives a set of elements for the cone. Project 
these points on the circle to base fine P T' of elevation 
and thence to T". Where the lines in the boss cross 
these element lines, as for instance where line 2' 2" 
intersects the element at R, drop lines to intersect the 



boss. Take the distance 3 r 3 in the profile and place 
it both sides of X Y as 3* 2> XI . Do same with space 
2 T 2 and 1 T 1. A line through points Z , 1* etc., 
gives true profile of the boss. 

For the pattern draw a line U V at right angles to 
line o' o" and place thereon the spaces 3* around to 
2 ZX in the true profile as s to 6 s . Draw lines through 
these points parallel to o' o" and intersect them by 
lines projected from like points in the boss as shown 
in the diagram. A line traced through the points ob- 
tained by that procedure will be the pattern to which 
necessary laps are to be allowed. Exactly the same 
procedure is followed for the other boss as directed 
by Fig. 790. 



450 



The New Metal Worker Pattern Book. 

PROBLEM 231. 

Pattern for an Unusual Five-pronged Fork. 



Many and diverse are the problems arising in the 
trade and herewith is a demonstration how to lay out 
the patterns for a five-pronged fork shown in Fig. 791. 
The lower pipe is a rectangle, from each corner of which 
a fork is to project with a round outlet. Through the 
center another fork projects. 

All parts marked A indicate the round pipe in both 
plan and elevation. As the corner forks have equal 




Fig. 791 . — The Five-pronged Fork. 

flare, the pattern for one can be used for all. In devel- 
oping this corner fork only one-quarter plan will be 
required as indicated by a b c. 

First draw any center line as A B on which estab- 
lish any point as C, and from this point draw the one- 
quarter rectangle as indicated by C 1 1° 12. From 1° 
draw the desired angle of the fork as shown by 1° a. 
Establish the length of the fork on its center line as 
15, through which draw the vertical line 13 17, repre- 
senting the top line of the round pipe in plan. Make 
15 13 and 15 17 each equal to the semi-diameter of the 
pipe, and draw a line from 17 to 1 and 1° and from 13 to 
1° and 12. 

From 15 draw the horizontal line 15 15' indefinitely, 
on which establish the center m, with which describe 
the circle G' equal to 13 17 as shown. Divide this 
circle into equal spaces from 13' to 20', through which 
points draw horizontal lines, cutting the line 13 17 and 
from these points draw lines to the apex 1°. 

With C as center and the desired radius, draw the 



quadrant 5 8, which divide into equal parts as shown by 
5, 6, 7 and 8. This quarter circle 5 8 represents the 
quarter section along the base of the center prong 8 5 in 
elevation. 

The quarter circle no in plan shows the quarter 
section on the top line of the prong D E in elevation. 
Through any point as 12 in elevation, draw the hori- 
zontal line X Y. Establish at pleasure the liight of 
the prong as 12 8 in elevation, and draw the horizontal 
line 8 7 6 5 representing the base line of the center prong- 
as shown by the dotted lines projected from the plan. 
Project the point 1 in plan as shown by 1 in elevation 
and at pleasure draw any desired profile as 5 1. Divide 
this into equal spaces shown by 1, 2, 3, 4 and 5 and drop 
these points in the plan as shown by similar numbers. 

A true section must be drawn at pleasure on 8 12 
in plan as follows: At right angle to 8 12 in plan from 
the point 8 erect the line 8 8° equal in liight to 12 8 
in elevation, and from the point 8° in plan draw at pleas- 
ure the profile 8° 12, which divide into equal spaces 
as shown by 8°, 9, 10, 11 and 12, from which points 
draw horizontal lines cutting the line C B in plan at 8, 
9, 10, 11 and 12 and vertical lines cutting the base lines 
X Y in elevation as shown by similar numbers. 

Using 12 in elevation as center and with radii equal 
to the various numbers draw quadrants intersecting the 
vertical line A B at 8, 9, 10 and 11. From 12 in eleva- 
tion draw the desired angle of the center of the prong 
as indicated by 12 b, which intersect at 13 17 by a line 
erected from similar numbers in plan as shown. From 
13 17 in elevation draw a horizontal line, on which estab- 
lish any point as s which use as a center and describe 
the circle G similar to G 1 . Divide G into the same 
number of parts as G 1 , being careful that if 13' 17' runs 
vertical in G 1 it will run horizontal in G. From the 
various points in G project lines to the line 15 19. Draw 
lines from 15 to 8 and 19 to 1 and draw the center prong 
D E 5 8 to correspond to the plan as shown. 

G and G 1 then represent the section of the top of 
the prong on the lines 15 19 in elevation and 13 17 in 
plan respectively; 1 5 in elevation the true section on 
the line 1 5 in plan; the quadrant 5 8 in plan the true 
section on the line 5 8 in elevation; the section 8° 12 
in plan the true section on the line 8 12 in plan or eleva- 
tion and the quarter rectangle C 1 1° 12 in plan the 
section on the base line 12 1 in elevation. 



Pattern Problems. 



451 



To avoid a confusion of lines separate plans have 
been drawn showing the base lines in the bottom and top 
of the prong. Thus 1 1° 12 13 
17 shows the one-quarter plan 
of the base lines in the bottom 
of the prong, while A x , which is 
a reproduction of B x , shows how 
the base lines are drawn in the 
top of the prong. 

Note in the plan B 1 or bot- 
tom the side 11° connects to 17, 
while the side 12 1° connects to 
13 and all the spaces on 13 17 
are connected to 1°. In the 
plan A x or top, points 1 to 5 
connect to 17; points 17 to 15 
to 5; points 15 to 13 to 8; 
points 5 to 8 to 15 and points 8 
to 12 to 13. 

Parallel to the base line X 
Y from the various intersections 
in the elevation from 1 to 5; 
5 to 8, 8 to 12 and 13 to 20 
draw lines to the right indefi- 
nitely, these hights representing 
the altitudes of the various tri- 
angles which will be constructed 
with bases equal to similar 
numbered lines in the quarter 
plans B x and A 1 . 

To obtain the true lengths 
of the base lines in the bottom 
of the prong B x take the various 
lengths of the lines in B x and 
place them as shown on the line 
X Y in elevation. From these 
points erect perpendiculars in- 
tersecting similar numbered 
horizontal lines previously drawn 
from similar numbered points in 
the elevation. 

For example: To find the 
true length of the hne 12 13 in 
plan take this distance and 
place it as shown from 12' to 
13' on the line X Y. As 12 in 
elevation is on the base hne, 

then 12' remains in the position shown, but from 13' 
erect the perpendicular line, intersecting the horizon- 
tal line drawn from 13 in elevation at 13. Then 13 12' 



is the true length of the hne shown by 13 12 in 
plan. 




Pattern for Corner Prong 
All FburAhhe 



Pattern for Center Prone 



Fig. 792. — Details of Pattern for Five-pronged Fork. 



To avoid a confusion of lines in finding the true lengths 
of the fines in the top, a reproduction of the one-half 
elevation showing the various altitudes has been made in 



452 



The New Metal Worker Pattern Book. 



C x with all the required numbers and intersections. 
From the various points in (T draw horizontal lines to 
the right. Extend the line 12 1 in C 1 indefinitely to the 
right as shown, on which place the various lengths shown 
in the quarter plan A* as shown by similar numbers on 
the lines 12 T, 8 V and 15 W in C x . At right angles 
to these lines from the various intersections on same 
draw lines intersecting similar numbered horizontal 
lines from the various numbers in the elevation C z . 
Connect points by lines as shown, which will be 
the true lengths of similar numbered lines in the 
plan A 1 . 

Thus to find the true length of the line 9 13 in plan 
A 1 set off this distance as shown from 9' to 13' on the 
line 12 T. From 9' and 13' erect perpendiculars inter- 
secting horizontal lines drawn from 9 and 13 in C J , 
thus obtaining the intersections 9 and 13, which be- 
comes the true length of 9 13 in A r . 

The pattern may now be laid out as follows: If 
the seam is to come on 1 19 in elevation or 1° 19 in plan, 
take the true length of 1° 19 in the true lengths in K and 
place it in the pattern in M as shown from 1° to 19. 
Using 1° in M as center and a radius equal to 1° IS 
and then 17 in K, draw the arcs 18 and 17 in M. Set 
the dividers equal to 19' to 18' to 17' in G 1 and start- 
ing from 19 in the pattern M step to arc 18 then to 17 
and draw a line from 17 to 1°. With 1° in 1 plan B x 
as radius and 1° in M as center draw the arc 1, which 
intersect by an arc struck with 17 as center and 17 1' 
in K as radius. Draw a line from 17 to 1 in M. With 
radii equal to 17 to 2, to 3, to 4, to 5 in L and 17 in M 
as center draw the arcs 2, 3, 4 and 5. Set the dividers 
equal to the spaces between 1 and 5 in the section H in 
elevation and starting from 1 in M step to arc 2, 3, 4 and 5 
and draw a line from 5 to 17. 

Again referring to diagram L obtain the radii 5 to 15 
and 16 and using 5 in M as center describe the arcs 16 and 



15. Set the dividers equal to the spaces between 17' 
and 15' in either G or G 1 , and starting from 17 in M 
step to arc 16 then to 15 and draw a line from 15 to 5. 
With radii equal to 15' 6, 7 and 8 in L and 15 in M as 
center describe the arcs 6, 7 and 8 which intersect by 
divisions obtained from 6, 7 and 8 in the quadrant in 
plan B x , and draw a line from 8 to 15. With radh equal 
to 8 14 and 13 in L and 8 in M as center draw the arcs 

14 and 13 and intersect same by spaces obtained between 

15 and 13 in G 1 . Draw a line from 13 to 8. 

Again using the radii in L from 13 to 9, 10, 11 and 
12' with 13 in M as center draw the arcs 9, 10, 11 and 12, 
which intersect by arcs equal to the spaces in the profile 
J from 8° to 12 in plan. Draw a line from 12 to 13 in 
M. With 12 1° in plan as radius and 12 in M as center 
describe the arc 1°, which intersect by an arc struck 
from 13 in M as center and 13 1° in K as radius. Draw 
a line from 1° to 13 in M. With radii equal to 1° 20 
and 19 in K and 1° in M as center describe the arcs 20 
and 19, which intersect by arcs with spaces equal to 
13 to 19 in G or G 1 . Draw a line from 1° to 19 in M and 
trace a line through the various intersections obtained. 
Then will 1° 1 5 8 12 1° 19 13 17 19 be the pattern for 
the corner fork, four of which will be required, two formed 
right and the other two left. 

The pattern for the center fork is obtained by extend- 
ing the hue 5 E in elevation until it intersects the center 
line A B at F. Then using F E and F 5 as radh and F° 
in diagram N as center describe the arcs E 1 E° and 5 5°. 
From any point as 5 draw a radial line to F° intersecting 
the inner arc at E 1 . On the outer arc 5 5° lay off the 
girth of four times the number of spaces contained in 
the quarter circle 5 8 in plan as shown from 5 to 8 to 8' 
to 8" to 5° in N. From 5° draw a radial line to F° and 
5 5° E° E 1 is the pattern for the center fork. 

Allowance must be made on all patterns for riveting 
or seaming. 



PROBLEM 232. 



Circle to Ellipse Offset Transition Piece. 



An instructive demonstration of the method for 
developing patterns for a square elbow having an ellip- 
tical base 7X11 inches to an 8-inch circular section 
offsetting at an angle of 30 degrees when viewed in 
plan, as shown by Fig. 793. 

In developing these patterns the elliptical pipe can 
be developed by parallel lines while the circular pipe 



must be laid out by triangulation. The first step is to 
draw the plan of the elbow as shown in the accompany- 
ing illustration, in which 8 2 4 6 shows the plan of the 
elhptical pipe. Draw the two diameters in the ellipse 
which intersect each other at V. From this intersection 
V draw a hne at the desired angle or 30 degrees, as shown 
by V 1'. 



Pattern Problems. 



453 



Establish the desired length of the arm of the elbow 
from V to 1' and at right angles to this line, through 
the point 1' draw the line 3' 7', making 1' 7' and 1' 3' 
equal to the semi-diameter of the S-inch circular pipe. 
In line with 3' 7' in plan draw the profile of the circular 
pipe, as indicated by X, as shown. This completes the 
plan view of the elbow, offsetting at an angle of 30 
degrees. 

Before the side elevation of the elbow can be drawn 
(which however is not necessary in the development of 
the patterns), an oblique elevation must be drawn as 
follows: At right angles to V 1' in plan, draw the oblique 
elevation of the elbow, as shown by 5' 5 S T 1 1, mak- 
ing both arms of the elbow at right angles to each other, 
both in the throat and on the heel, as per requirements 
of the problem. In line with 1' 5' in the oblique view 
draw the profile Z similar to the profile X in plan. 

Assuming that the seams are to come in the throat 
of the elbow, start point 1 in the throat line in profile Z 
and divide the profile in an equal number of divisions, in 
this case 8, as shown from 1 to 8 to 1. From these small 
figures parallel to the arm 5' 5 in the oblique view draw 
lines to the left, cutting the elevation line from 1' to 5', 
as shown. In a similar manner divide the profile X in 
plan in similar number of spaces, as shown from 1 to 8 
to 1, being careful to start 1 in the profile X, as indicated, 
thus making a quarter turn from that in the profile Z. 

From the small figures 1 to 8 in X, parallel to the 
30 degree fine V 1' draw lines as shown, cutting the 
plan fine from 3' to 7', as shown. 

Now divide the elliptical profile in an equal number 
of divisions, being careful to have the points 1 and 5 
at the points of tangency where lines drawn at right 
angles to V 1' are tangent to the extreme outline of the 
elliptical section, as indicated by 1 and 5. Note that 
the two ends of the ellipse have been divided into smaller 
spaces than the sides, all as indicated. 

In developing the transition arm from circular to 
elliptical, triangulation will be used, and as the circular 
section is divided into 8 parts and the elliptical section 
into 12, for that reason are the divisions in the elliptical 
section numbered 1 1°, 2, 3 3°, 4, 5 5°, 6, 7 7°, 8, so that 
the two ones, the two threes, the two fives and the two 
sevens will join with 1', 3', 5' and 7' respectively when 
joining the base fines of the triangles in plan. 

As previously mentioned, the side elevation is not 
necessary in the development of the patterns and if it 
was desired to draw this correctly, inasmuch as it is 
good practice and allows of a better conception 
of the problem, it would be done in the following 



way: From the small figures in the elliptical profile in 
plan, drop lines at right angles to V 1' crossing the miter 
line 5 1 in the oblique elevation, as shown by similar 
numbers and intersecting the base line of the elbow 
S T, as shown. 

In similar manner at right angles to the diameter 4 8 
in the elliptical profile in plan, erect fines indefinitely and 
at pleasure draw the line S° T° parallel to 4 8 in plan. 

Now, measuring in each instance from the line 
S T in the oblique elevation, take the various distances 
to the numbered points on the miter line 5 1 and place 
them on similar numbered fines in the side elevation, 
measuring in every instance from the line S° T°, thus 
obtaining the points of intersection shown, which repre- 
sents the miter line between the two arms. 

From the intersections 1 and 5 which represent 
respectively the throat and heel fines of the elbow, draw 
horizontal fines to the right, which intersect by the 
vertical fine erected from the point 1' 5' in plan, thus 
obtaining points 1" and 5". In line with 1" 5" in the 
side elevation draw the profile Y similar to profiles X or 
Z. As point 1 begins at tfie throat fine in profile Z in 
the obhque elevation, then, also begin 1 in the throat line 
in profile Y in the side elevation, as shown, and space the 
profile in eight equal spaces. From these small figures 
2 to 8 draw horizontal lines to the left and intersect same 
by vertical lines erected from similar numbered points 
in plan, thus obtaining the points of intersection 8" 
to 2" in the side elevation. Then will T° 1 1" 5" 5 S° 
be the true side elevation of the elbow. 

The patterns are now in order and the pattern for 
the elliptical pipe is obtained as follows: Extend the line 
T S in the oblique elevation as T l x and starting torn 
1° set off the girth of the elliptical profile in plan, as 
shown by similar numbers on T I s . At right angles to 
T l z from the small figures on same, draw lines which 
intersect by lines drawn parallel to T l x from similar 
numbered intersections on the miter line in the oblique 
elevation. A fine traced through points thus obtained, 
as shown by 1 1" l x 1 will be the desired pattern. 

Now connect the base lines in plan from which the 
true lengths are obtained. Draw solid fines from 1 and 
1° to 1'; from 2 to 2'; 3 and 3° to 3'; 4 to 4'; 5 and 5° 
to 5'; 6 to 6'; 7 and 7° to 7' and 8 to 8'. Draw dotted 
lines from 1° to 2'; 2 to 3; 3° to 4'; 4 to 5'; 6 to 5'; 
7 to 6'; 8 to 7' and 1 to 8'. These lines then represent 
the bases of triangles which will be constructed whose 
altitudes will equal the difference between similar num- 
bered points in the oblique elevation. For example: 
To find the true length of the solid fine 2 2' in plan, take 



454 



The New Metal Worker Pattern Book. 



this distance and set it on the horizontal line 2 2' in the 
true solid length. 

5 Side Elevation 5 



the miter line carry lines to the right, meeting the line 
a b, as shown. In same way from intersections on the 




True Lengths of Dotted Lines 
Fig. 793. — Patterns for Offsetting Transition Elbow. 



—5' True Lengths of 
Solid Lines 



At right angles to 2 2' erect a line from point 2 
equal to the difference in hight between 2 and 2' in the 
oblique elevation. These hights are found as follows: 
At pleasure, at right angles to the upper arm of the elbow 
draw any line as a b. From the various intersections on 



line 5' 1' in the oblique elevation carry lines to left, cut- 
ting the line a b. 

Now find the difference in hight between 2 and 2' 
by following the line drawn from 2 and 2' to the line a b 
indicated by h i. Set off this hight h i from 2 to 2", 



Pattern Problems. 



455 



as shown in the true lengths. Then will the slant line 
marked T L from 2" to 2' be the desired length. In 
similar manner obtain all of the true lengths on solid lines 
in plan. The true lengths of the dotted lines are ob- 
tained in the same manner. For example take the length 
of the dotted line 6 5' in plan and place it as shown by 
6 5' in the true dotted lengths. 

Find the difference in hight between 6 and 5' by 
referring to the line a & in the oblique view where the 
height is indicated by n o, which is placed as shown by 
6 &. &" 5' is then the desired true length. As there 
is no difference in hights between 1 1' and 5 5' in the 
oblique elevation then will 1 1' and 5 5' in plan show 
then - true lengths. 

When laying out the pattern for the transition arm, 
the true lengths just obtained, the divisions in the circles 
X, Y or Z and the divisions along the miter cut in the 
elliptical pipe pattern are used. 

The pattern may now be laid out. Take the dis- 
tance of 1 1' in plan, which shows its true length and place 
it as shown by 11' in the pattern P. Now with the 
radius equal to 1 1° in the miter cut in the pattern for 



elliptical pipe and with 1 in the pattern P as center 
describe the arc 1°, which intersect by an arc, struck 
from 1' as center, with the (T. L.) of the solid line 1' 1° 
in the true lengths of solid lines as radius. Now with 

1 2 in either circular profiles as radius, and 1' in the pat- 
tern P as center, describe the arc 2' which intersect by 
an arc struck from 1° as center, and the (T. L.) of the 
dotted line 1° 2' in the true lengths of dotted lines as 
radius. Now with 1° 2 in the miter cut in the pattern 
for elliptical pipe as radius, and 1° in the pattern P as 
center describe the arc 2, which intersect by an arc, 
struck from 2' as center and the (T. L.) of the solid line 

2 2' in the true lengths of solid lines as radius. 

Proceed in tins manner, using alternately, first the 
divisions in the circle, then the proper true length; the 
divisions in the miter cut in the elliptical pipe pattern, 
then again the proper true length until all the spaces 
have been transferred. The length 5 5' in pattern 
is obtained from 5 5' in plan. Trace a line through 
points thus obtained in P. Then will 1 5 1 1' 1' be 
the desired pattern. Allow edges for seaming or 
riveting. 



PROBLEM 233. 



Pattern for Fitting of Unusual Design. 



The irregular fitting shown in Fig. 794 presents some 
interesting problems in pattern development. The lower 
pipe is 24 inches square, the upper is a 36-inch octagon, 
in section, leading to a 36-inch round pipe. Fig. 795 
shows the developing the patterns. First, draw the side 
view of the fitting according to the dimensions and pro- 
jections desired, and indicated in the side view by 1, 8, 
8 1. In its proper position below 8 8 draw the profile 
of the square end of the pipe, and above 1 1 the profile 
of the octagon. Project the corners of the octagon to 
the line 1 1 in the side view indicated by 14 14. From 
these two points of intersection draw graceful curves, 
gradually tapering to the corners, 8 and 8 and 7 and 7. 
Great care must be taken in drawing these curves, or 
miter fines, in the side view, because when once drawn 
they become fixed lines which cannot be changed in 
developing the patterns. This, then, completes the side 
view, showing the transitions D and E from octagon to 
square. 

To the right, in line with the side view, draw the 
front view of the fitting. Both halves in this view are 
symmetrical on either side of the center line. In order 



to avoid drawing the octagon and square profiles in the 
front view, measurements are taken from 1 to 14 in 



36'Round 




Side 



Front 



Fig. 794- — Front and Side View of Fitting. 

the side view, and placed, as shown, from 14 to 1 in the 
front view, on either side of the center line. From 
point 7 in the side view, where the miter line ends, project 



456 



The New Metal Worker Pattern Book. 



this point horizontally to the right, cutting the outline 
of the front view also at 7. Now from point 1, in the 
front view, draw a gradual tapering double curve to 



Pattern'] 
torS.des; 

C-C in ! 

Profile , 




Pattern 
forTransition' 
Shown by D-D 
in Profi 




Pattern for 
Side B m 
Profile 




Pattern for 
Transition 
I* shown byE-E 
in Profile 



Front View 
H 



Pattern 
for Side A 
in Profile 



True Lengths True Lengths 
W of Transit.ons of Transitions 
D'm Profile E in Profile 




True 
Lenqth I 
ofa-2 
m Plan 



Diagram 



Fig. 795. — Front and Side 



Views, Profiles, True Lengths, and the Various 
Pattern Shapes. 



point 7. This can be traced to the opposite side of the 
center line, although one-half of the front view is all that 
is required, because both halves are alike. This then 
completes the front view showing the miter or joint 
line of the object or fitting. 



The four sides of the fitting indicated by A B C C in 
the profile can be laid out by parallel lines, while the 
sides D D and E E in the profile, which form the tran- 
sition from octagon to square, must be laid 
out by triangulation. To avoid a confusion 
of lines only one outline (the left in the side 
view) will be equally divided and from these 
points horizontal hues are carried through the 
two views, cutting the right-hand outline in 
the side view and the left-hand outline in 
the front view. This will divide these out- 
lines in unequal parts and also intersect the 
various miter lines, all as shown. The first 
pattern to be developed will be that shown 
by A in the profile. Therefore, extend the 
center line in the front view, H J, on which 
place the girth of the outline shown from 1 
to 8 on the left side of the side view, as 
indicated by similar figures 1' to 8' on H J. 
Through these small figures at right angles 
to H J, draw measuring lines indefinitely, in- 
tersecting them by lines drawn parallel to 
H J from similarly numbered intersections 
on the miter line in the front view. Trans- 
fer these points of intersection on the oppo- 
site side of the center line H J, and trace a 
line through the points of intersection thus 
obtained. This forms the pattern for side A 
in the octagon profile. Now extend the cen- 
ter line upward in the front view as in- 
dicated by G F. On this place the girth of 
the outline on the right side in the side view, 
measuring each space separately as they are 
all unequal as indicated from 1' to 8' on F G. 
Through these small figures draw perpendic- 
ulars to F G, intersecting them by lines drawn 
parallel to F G from similar points in the right 
side miter line in the front view. Transfer 
One these points in the pattern on the opposite 
side of F G, and trace a fine through the 
various intersections, which becomes the pat- 
tern for the side marked B in the octagon 
profile. As both curves in the front view are 
alike, the patterns for the two sides marked C 
and C in the octagon profile will be similar. To 
obtain this pattern erect any line at right angles to 1 1 in 
the side view shown by K L, and on this place the girth 
of the curve, 14 8, in the front view, measuring each space 
separately, as indicated by the small figures 8" to 14" 
on K L. Through these points at right angles to K L 



Plan 



Pattern Problems. 



457 



draw lines to the left indefinitely, which are intersected 
by lines drawn parallel to K L from similarly numbered 
intersections on both miter lines in the side view. Trace 
a line through the points thus obtained, from 14 to 8 to 
8' to 14', which will be the pattern for the two sides C C 
in the octagon profile. The next patterns in order are 
the two chamfers or transitions, D and E, in the side 
view. As the true lengths of the edge lines along 1 7 
and 7 14 in D, and 14 7 and 7 1 in E in the side view 
are obtained from the miter cuts in the patterns for A C 
and B respectively, then the next step is to find the true 
lengths of the connecting fines in both views as follows, 
Draw lines in both D and E in the side view from 2 to 14, 
3 to 13, 4 to 12, 5 to 11 and 6 to 10; in a similar manner 
connect similar lines in the front view. Care should be 
taken to number the points in both views accurately. 
Thus the outline 8 to 14 in front view represents the 
section through similar points in the side view. Out- 
lines 1 to 8 in the side view show sections through the 
center line in the front view. These various lines 1 
to 14 to 2 to 13 to 3, etc., in the front view, then rep- 
resent the bases of triangles which will be constructed, 
whose altitudes will equal the horizontal distances be- 
tween similarly numbered points in the side view. They 
are constructed as follows: Below the side view draw 
any vertical fine as M N, upon which place the various 
distances shown in the front view by 1 14, 14 2, 2 13, 
13 3, etc., until the length 6 9 is reached as developed 
on the line M N from 1' to 9'. From these various 
small figures at right angles to M N, draw lines indefin- 
itely, intersecting them by fines drawn parallel to M N 
from similarly numbered intersections in D and E in 
the side view. Connect the points in the diagram of 
true lengths, shown by the heavy lines, those to the left 
indicating the true lengths of similarly numbered lines 
in D, in the side view, and those to the right indicating 
the true lengths of similarly numbered lines in E in the 
side view. In developing the pattern for the transition 
piece D, the true lengths for D are used in connection 
with the divisions along the miter cut in the pattern for 
A, and the divisions along the left miter cut in the pat- 
tern for C C. When developing the pattern for the 
transition, E, the true lengths for E are used in con- 
nection with the divisions along the miter cut in the 
pattern for B, and the divisions along the right miter cut 
in the pattern for C C. 

For the pattern for the transition shown by D in 
side view and profile, proceed as follows: Take the length 
of the side D in profile or 1 14 in the true lengths D and 
place it as shown by 1 14 in B. Now, with 1 2 in the 



miter cut in the pattern for side A as radius, and 1 in R 
as center, describe the arc 2, which intersects an arc, 
struck from 14 as center and 14 2 in the true lengths for 
D as radius. Now, with 14 13 in the left miter cut in 
the pattern, C C as radius, and 14 in R as center, describe 
the arc 13, which intersects an arc, struck from 2 as a 
center and 2 13 in the true length for D as radius. Pro- 
ceed in this manner, using alternately, first the proper 
divisions along the miter cut in pattern for side A, then 
the proper true length in D; the proper division along 
the left miter cut in the pattern for side, C C, and again 
the proper length in the true lengths for D, until the line 
6 9 in R has been obtained. Now using 6 7 in the pat- 
tern for a side, A, as radius and 6 in R as center describe 
the arc 7, which intersects an arc struck from 9 as center 
and 9 7 in the left miter cut in the pattern for a side and 
C C as radius. Trace a line through the points thus 
obtained in R, and this will be the desired pattern. 

In a similar manner obtain the pattern for the transi- 
tion side marked E in side view and profile. Take the 
length of the side E in the profile or 1 14 in the true 
lengths for E and place it as shown by 1 14' in S. Now, 
using 1 2 in the miter cut in the pattern for side B as 
radius and 1 in S as center, describe the arc 2, which will 
be intersected by an arc struck from 14' as center and 14 2 
in the true lengths for E as radius. Now, with 14' 13' 
m the right miter cut in the pattern for sides C C as 
radius, and with 14' in S as center, describe the arc 13' 
which will be intersected by an arc struck from 2 as a 
center, and 2 13 in the true lengths for E, as radius. 
Proceed in this manner using alternately, first the divisions 
along the miter cut in the pattern for side B, then the 
proper true length in E; the divisions along the right 
miter cut in the pattern for sides C C and again the 
proper true length in E, until the line 6 9' in S is 
obtained. Now, using 9 7' in the right miter cut in pat- 
terns for sides C C as radius and 9' in S as center describ- 
the arc 7, which will be intersected by an arc struck from 
6 as center and 6 7 in the miter cut in the pattern for side 
B as radius. Trace a line through the points thus ob- 
tained. This will be the desired pattern. 

Another pattern will be required, that of the transi- 
tion from octagon to round, as shown in Fig. 794. This is 
shown developed in diagram X in Fig. 795. First, draw 
the plan view of the octagon and circle, through the 
center of which draw the seam line, as shown. Con- 
struct the elevation above the plan, although this is not 
necessary in practice, all that is required being the ver- 
tical hight indicated by a a'. As each of the eight sides 
of the transition piece are similar, all that is necessary is 



458 



The New Metal Worker Pattern Book. 



to draw radial lines from the corners a a in plan to the 
center e, intersecting the circle at 2 and 2. Divide the 
distance between 2 and 2 in equal parts (in this case two) 
as indicated by 1. As the side of the octagon is tangent 
to the circle at 1, the true length on this point is equal to 
the vertical hight of the article indicated by a a' and the 
true length of a 2 in plan is found by placing this dis- 
tance from a' to 2 perpendicular to a a', when a 2 will be 
the desired length. 

The pattern can now be laid out as shown above the 
elevation, as follows: Take the vertical hight of the 
transition piece a a' and place it in the pattern shown by 
1 1'. From 1' draw the perpendicular line 1' a equal to 
one-half the octagon side indicated by 1 a in the plan. 
Draw a line from a to 1 in the pattern. Now using a 2 



in the true length as radius, and a in the pattern as center, 
draw the short arc at 2. Again using a as center and a 1 
as radius, continue a short arc opposite 1 at 1°. Now set 
the dividers equal to the spaces 1 2 in plan. Starting 
from 1 in the pattern, step to arc 2 and 1° and draw line 
from 1° to a. Using 1° as center and 1° a as radius, 
draw the short arc a° . Now take the full side a a in 
plan and set it off from a to a° in the pattern. Draw 
line from a° to 1° and continue in this manner until the 
opposite half side is obtained, as indicated by a 1" 1. 
Trace a line through points thus obtained, then 1 1' 
and 1" 1 will be the half pattern for the transition 
piece octagon to round. All patterns shown are net, 
to which edges "or laps must be allowed for seaming and 
riveting. 



PROBLEM 234. 



Pattern for French Style Auto Hood. 



A very interesting demonstration" follows on how to 
develop the patterns for a French style of motor hood 
as shown in the sketch, reproduced in the accompany- 
ing illustration by the plan, side and rear views, Fig. 796. 
As the corners of the base of the hood are squares, that 
is, the sides at the base having no taper, as shown by 
C 1 C in plan, and as the top of the hood shown by 
E° 7 7" F° has a taper, thus making D° C X 6 in plan 
a warped surface, for this reason the sides will have to be 
developed by triangulation. The miter lines of the 
chamfer both in plan and elevation have been drawn at 
pleasure and can remain so, with the exception of the miter 
line 1 to 3 in plan, which line must be projected from 
the side view so that the lower mould of the base shown 
from 1 to 3 in the rear view can be developed by parallel 
lines. The first step is to draw the plan of the rectan- 
gular base as shown by H 1" 1 C, above which draw the 
rear of the hood as shown by H G F E D C, drawing 
the curves and angles to any required dimensions. From 
the corners of the chamfer G F and E D in rear view, 
project lines in the plan, thus obtaining G° F° and E° D°. 
Now from the corner C in plan erect the vertical line C E 1 
and from points E and D in rear view draw horizontal 
lines intersecting the vertical lines previously drawn at 
E 1 and D 1 . Then, with C as center and radii equal to 
C D 1 and C E 1 draw arcs intersecting the horizontal line 
drawn from C in plan at P 2 and E 2 . Draw the desired 
pitch E 2 7 in side view and parallel to it, the line D 2 6. 



Extend the line 7 6 in side view cutting the plan at the 
desired locations as 7 and 6 respectively. Now from 
the points 7 and 6 in the side view draw to the required 
dimensions graceful curves from 7 to 1 and 6 to 1. As 
the lower part of the side of the base shown from C to 3 
in rear view will be developed by parallel lines, the true 
miter line in plan must first be found before the entire 
miter line of the chamfer can be drawn in the plan view. 
This is accomplished as follows: Divide the small curve 
in rear view into equal spaces as shown from 1 to 3, from 
which points horizontal fines are carried to the right, 
cutting the vertical line C E 1 as shown by the heavy dots. 
Using C as center with these various dots as radii describe 
arcs cutting the perpendicular line in side view at 1 2 
and 3 as shown. Parallel to the base line 1 1 in side 
view, lines are drawn from the points 1, 2 and 3 on 1 E 2 
until they cut the inner curve of the chamfer in side view 
also at 1, 2 and 3 and from which points they are drawn 
horizontally or parallel to 1 1° in plan and are intersected 
by vertical lines dropped from 1, 2 and 3 in the rear view, 
thus obtaining the points of intersections 1, 2 and 3 in 
plan. Now, through these points 1, 2 and 3 in plan 
draw a graceful curve until it meets the point 6 in 
plan previously obtained, which represents the plan view 
of the inner curve of the chamfer. From point 7 in plan 
draw a gradual graceful taper to point 1 as shown by 
the curve 7 1, which shows the plan view of the outer 
curve of the chamfer. The proper drafting of the rear, 



Pattern Problems. 



459 



side and plan views is half the battle. 
The pattern for the combined top and 
front piece is developed as follows : From 
the points 2 and 3 previously obtained 
in the side view draw horizontal lines 
cutting the upper curve in the side view 
at 11 and 10, and the inner and outer 
curves in plan view at 2, 3 and 11, 10 
respectively. Now, between the points 

7 and 10 a in the side view introduce 
equal spaces shown by 8 and 9 and in- 
troduce an extra point between 7 and 

8 where the curve is small as at a'. 
From these various intersections 7 to 9 
in" side view draw lines indefinitely in 
the plan, cutting the inner curve in 
side view at 6, a, 5 and 4, and corre- 
sponding points in the plan as shown by 
similar numbers and letters. Now ex- 
tend the center fine A B of the hood as 
shown by A J, upon which place the 
girth of the top and front of the hood 
shown in the side view by E 2 7 a' 8 9 
10 11 1 C and as shown by similar 
numbers on B J. At right angles to 
B J through these small figures draw 
fines as shown which intersect by ver- 
tical fines, partly shown, drawn parallel 
to B J from similar numbered intersec- 
tions in plan. Trace a line through 
points thus obtained as shown by the 
intersections E" to C" and trace this 
cut opposite the center line B J, which 
will show the full pattern for the top 
and front of the hood, and the spaces 
along the miter cut E" C" will give the 
true top edge line when developing the 
chamfer pattern. The pattern for the 
lower part of the side is developed by 
extending C H in plan as H P upon 
which place the girth of the mold C to 
3 in rear view as shown by the small 
figures C, 1, 2, 3 on HP. From these 
small figures perpendiculars are drawn 
and intersected by fines drawn parallel 
to H P from similar numbered points 
C, 1, 2, and 3 in plan view, as shown. 
Trace the miter cut 1" to 3" in the 
pattern for sides. Then, 3", 3 C 1" is 
the pattern for the lower curve to 



Rear View 
,A E 




Fig. 796. — Pattern for French Auto Hood. 



460 



The New Metal Worker Pattern Book. 



which the balance of the side piece must be added 
by triangulation. Where the line 7 6 extended in side 
view, crosses the lines 3 in both side view and plan, 
designate these two points X, as shown. Now, from the 
intersections 4, 5, a, 6 and D° in plan, draw lines to X and 
in a similar manner connect similar points to X in the 
side view as shown. The lines in plan will then rep- 
resent the bases of triangles which will be constructed, 
whose altitudes will equal similar numbered hights in 
the side view and are obtained as follows: Extend the 
line 3 X in side view, as shown by X Y, upon which place 
the various lengths of X D°, X 6, X a, X 5 and X 4 in 
plan as shown by similar numbers on the line X Y. 
From these points at right angles to X Y, erect lines 
intersecting them with lines drawn from similar num- 
bered points in the side view, thus obtaining the points 
of intersections D" 6", a", 5" and 4". 

From these points draw lines to X which show the 
true lengths of similar numbered lines in plan and side 
view. The true edge line must now be found along the 
miter line D° 3 in plan as follows: Draw any vertical line 
as L M, upon which place the girth of the miter line 
D°, 3 in plan measuring each space separately as shown 
by similar numbers on L M. Through these small fig- 
ures at right angles to L M draw lines making them 
equal in hight to similar numbered points, measuring 
in each instance from the line 3 3 in the side view and 
thus obtaining the points of intersections D" 6" a'" 5" 
4" and 3" in the developed section. A line traced through 
these points gives the developed section or true edge line 
along D° 3 in plan. The pattern is developed with 3 D 
in rear view as radius, and 3 in the pattern for sides S as 
center describe the arc D". Take the distance from 3 to X 
in the side view and place it as shown from 3 to X in S. 
Now using the true length X D" in T as radius and X in 
S as center intersect the arc previously struck at D". 
Now, with radii equal to X 6", X a", 5" and 4° in the 
true lengths in T, and X in S as center describe the short 
arcs 6", a", 5", and 4." Now set the dividers equal to 
the various spaces in the developed section V as 3" to 4" 
to 5" to a'" to 6" to D" and starting from 3" in S step 
to similar numbered arcs as shown. A line traced through 
points thus obtained as shown by D", 6", 1", C will 



represent the outline of the pattern for the sides. As 
the pattern for the chamfer will be considered next, the 
true lengths of the solid and dotted lines shown in the 
plan of chamfer must be obtained. This is accomplished 
as follows: Connect opposite points in both plan and side 
view of the chamfer as indicated from 6 to a', a to 8, 5 to 
9, 4 to 10, and 3 to 11. Now draw any line below the 
side view, as shown by N R, upon which place the various 
'lengths in plan as D° 7, 7 6, 6 to a' to a to 8, to 5, to 9, to 
4, to 10 to 3, to 11 to 2 and place them on the line N R, as 
shown by similar letters and numbers. From these 
points at right angles to N R, erect lines which intersect 
by lines drawn from similar numbered points in the side 
view, parallel to N R, thus locating the points D x , 7 X , 
6', a' x , <f, 8 X , 5 X , 9", 4*, 10*, 3 s , 11", 2 T , which points, 
when connected, show the true lengths of similar num- 
bered lines in plan and side view. The pattern for the 
chamfer shown by Z can now be developed as follows: 
Take the distance of D" 6" in S and place it as shown 
by D" 6" in Z. Now with 7 X 6 1 in the true lengths in U 
as radius and 6" in Z as center describe the arc 7", which 
intersect by an arc struck from D" as center and ~D X 7 X 
in U as radius. Now using D E in rear view as radius 
and D" in Z as center, describe the arc E" which inter- 
sect by an arc struck from 7" as center and 7" E" in the 
top pattern W as radius. Now with 7" a" in W as radius 
and 7" in Z as center, describe the arc a" which intersect 
by an arc struck from 6" as center and 6* a' x as radius. 
Then using 6" a" in S as radius and 6" in Z as center 
describe the arc a'", which is intersected by an arc struck 
from a" as center and a' x a x in U as radius. Proceed in 
this manner, using alternately first the divisions along 
the miter cut in W, then the proper true length in U; 
the divisions along the miter cut in S, then again the 
proper true length in U, until the last two points 11" 
and 2" in Z have been obtained. Then using 11" 
1" in W and 2" 1" in S as radii, and with the points 
11" and 2" respectively in Z as centers, describe arcs 
intersecting each other at 1". Trace a line through 
points thus obtained as shown by 1", 7", E", D," 6", 
1", which will be the pattern for the chamfers. Edges 
and laps must be allowed for wiring, seaming and 
riveting. 



Pattern Problems. 



461 



PROBLEM 235. 



Pattern for French Style Auto Hood. 



This style of hood is a very ingenious design. In 
the development of the pattern it is assumed that it 
contains part of a true cone located in each corner. A 
ribbon of uniform width runs from the dashboard to the 
cone and from the cone the ribbon runs down to the 



Draw one-half of the end elevation A as taken 
from the dashboard. Draw one-half of the plan B. 
Begin beside elevation C by drawing the base and run a 
line up from each extreme end. Draw a line from the 
dashboard at the desired pitch and let it intersect at X 




Pattern for Sides 








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Fig. 797. — Pattern for French Style Auto Hood. 



molding at the base of the hood and comes to a point at 
the corner. The surface of the cone and the surface of the 
ribbon must join in plane without angles at the point of 
intersection. As the top and sides are both shown in a 
foreshortened view, a little preliminary work is neces- 
sary before the patterns are developed. The method is 
as follows: 



Draw the line P through the side elevation C and the end 
elevation A. This line is level with the lower surface of 
the cone where it intersects with the ribbon. The angle 
of the front corner of the plan B must be divided equal 
and the distance E equal to the width of the ribbon laid 
off one-half on either side parallel to an imaginary line of 
intersection as shown at E. Project the point X' in the 



462 



The New Metal Worker Pattern Book. 



side elevation C to the plan B and also project the X' 
from the side elevation to the end elevation A, then down 
to the plan B, locating the same point in the plan B at 
the point of the arrow. Project the point X' to the true 
side elevation F, locating X as shown, and also project 
the point X' from the side elevation C straight through 
the end view A and then to point X in the true elevation F. 
Locate the point I in true elevation F by projecting 
from end elevation A and plan B. Also carry the line P 
through A and around to establish the line P in the true 
elevation, as shown. Next, project the lines J and J' 
from the plan B to the true elevation F, as shown. Draw 
the hues J I, as shown in true elevation F, bisecting the 
angle, then draw the line through I to S at right angles 
with the line H J. With I as the center, describe the arc 
through S to the line P. This is now part of the true 
elevation and may be divided into equal parts. From 
the true elevation F the lines are projected as shown 
until they intersect with lines from the plan B, forming the 
outline for a cone at G. The points of the cone are run 
down to the plan from the true elevation F and from the 
true elevation also through G to the plan. The miter 
line in the plan B is drawn through the points of inter- 
section of these lines. Projection of the points from the 
true elevation F to the end elevation A and from the 
plan B to the end elevation A, as indicated by the arrows, 
establishes the points on the miter lines at A. Projecting 
from the end elevation A to the side elevation C and from 
the plan B to the side elevation C locates the points 
of the miter lines and also the true profile of the hood. 
The lines shown at M in the true elevation F are spaced as 



far apart as the distance M taken from ihe end eleva- 
tion A. To get the pattern for the ribbon, draw two lines 
parallel to the sloping top of the true elevation F. The 
distance to make them apart is taken from E in the plan 
B. Continue the fine from I through S far enough past 
the end of the pattern to support the center of the pattern 
for the surface of the part formed by the cone. The 
distances to step off on the pattern for the cone part 
are taken from the true elevation F. The stretch- 
out for the lower part or tongue of the ribbon is 
taken from the profile D, and the width to space the 
points is taken from the plan B. The end of the ribbon 
tongue as projected from H is not quite square. The 
pattern for the part of the side which occurs below 
the fine P is developed, as shown from the plan B and 
the part of side above the line P, is copied from the 
true elevation F. The pattern for the top and front 
may be projected from the end of either the end eleva- 
tion A, as shown, or it could be projected from the 
plan B. The stretchout for the side below the line T 
and for the top and front is taken from the true profile 
in the side elevation C. 

No laps are added in this drawing for attaching the 
parts together. Any desirable ornamentation, either 
raised or sunk, can be placed on the front and side panels 
by marking them on the pattern in the proper location 
before the parts are formed into shape. If they are 
added, make the proper laps for the depth of the panel 
and the metal required to secure them into place and 
leave the proper margin all around when cutting an 
opening for a sunken ornament. 



PROBLEM 236. 



An Automobile Fender of Special Design. 



An interesting feature of this problem in descriptive 
geometry is the complete set of full-size measurements 
given on the drawing repioduced in Fig. 798. 

All the measurements show in this illustration, with 
the exception of those given below the frame fine on the 
apron. These were omitted because they can be added 
to the pattern for splash guard or apron below the fine 
a 5', which represents the top of the frame fine. The 
first step is to correctly draw the side and end elevations 
of the fender as shown. While the plan is here shown, 
it is not necessary, excepting to know that the upper 
end of the top of the fender is a semi-circle, struck with a 



5-inch radius, being one-half of the 10-inch wide fender. 
Now, from the extreme upper end of the fender in eleva- 
tion, lay off one-half the width of the fender, or 5 inches, 
as shown from x to 1, and from 1 and the intersection 
between the fender and top frame line at 5 divide the 
curve in equal spaces, as shown from 1 to 5. Number the 
intersection between the fender and the bottom frame 
line, 6, and the extreme lower end of the fender, 7. As 
the Up of the rim will be seamed separately to the rim, 
fender and apron, therefore, from 1 in side elevation draw 
the vertical fine 1 0, and from draw the desired curve, 
a. Draw a line from 1 and to a, as shown. Now, 



Pattern Problems. 



463 



from the various intersections 1 to 7 in side elevation 
draw horizontal lines to the right, cutting the rear side 
of the fender in the end elevation also from 1 to 7. In 
similar manner, from points a, 5 and 6 in the side eleva- 
tion extend horizontal lines to the right, intersecting the 
frame or channel of the chassis in end elevation at a', 
5' and 6'. From 6' in the end elevation draw the desired 
curve at pleasure, at 6' to 7. Now, from points 1, 2, 3, 
and 4 in the side elevation draw lines to point 5. These 
various lines in the side elevation will represent the 
bases of triangles, which must now be con- 
structed, whose altitudes will equal the hori- 
zontal distances between similar numbered points 
shown in the end elevation, and are constructed 
as follows: Extend the lines a' 6' and 2, 7, in 
the end elevation indefinitely, as shown. Now 
take the various lengths, 5 4, 5 3, 5 2 and 5 1, in 
the side elevation, and place them in the dia- 
gram of true lengths below the end elevation, 
as shown by 5' 4', 5' 3', 5' 2', and 5' 1'. From 
the point 1' 4' draw horizontal fines to the left, 
intersecting the fine extended through 1 7 in 
end elevation at 1, 2, 3, and 4, in the true 
lengths. Now draw fines from 1, 2, 3, and 4 to 
5', which will show the true lengths of similar 
numbered lines in the end and side elevations. 
In similar manner take from the side elevation 
the length of the lines a and a 1 and place 
them in the true lengths, as shown, from a to 
0' and a to 1'. From 0' and 1' draw horizon- 
tal lines, cutting the line 2 7 extended at 0° 
and 1°. Draw lines from 0° and 1° to a, which 
are the true length of lines having similar 
numbers in either elevation. The pattern for 
the apron or splash guard can now be developed 
as follows: Take the distance of a 5 in side 
elevation and place it as shown by a 5' in S. 
Now, with a 1° in the true lengths as radius 
and a in S as center describe the arc 1, which inter- 
sect by an arc struck from 5' as center and 5' 1 in 
the true lengths as radius. Now, using radii equal to 
5' 2, 5' 3 and 5' 4 in the diagram of true lengths, and 5' 
in S as center, describe the short arcs near 2, 3, and 4. 
Now set the dividers equal to the spaces between 1 and 4 
in the side elevation and starting from 1 in the pattern 
S, intersect the arcs at 2, 3, and 4. Using 4 5 in the side 
elevation as radius and 4 in S as center, describe the arc 5, 
which intersect by an arc struck from 5' as center and 
5' 5 in the end elevation as radius. Now at right angles 
to 5 5' in S draw fines indefinitely from points 5 and 5' 



as shown. Take the distance from 5 to 6 and 6 to 7 in 
the side elevation and place it, as shown, from 5 to 6 and 
6 to 7 in the pattern S. In similar manner make the dis- 
tance 5' 6' equal to 5 6. As the curve 6' 7 in the end 
elevation has been established, this can be approximately 
transferred to the pattern as follows: Draw a line, as 
shown, from 6' to 7 in the end elevation, which bisect 
so as to find the depth of the curve at its center, as indi- 
cated by the arrows, and take this distance between the 
arrows, and set it off in the pattern S, from the center of 



Plan 




Pattern for Splash Guard or Apron 

Allow Flanges forSeamlng-Wiring or 
Bolting on all Patterns 



T 

i 



True Lengths of 

Similar Numbered 

Lines in Side or 

End Elevations 



Pattern, for Fender 
Fig. 798. — An Automobile Fender of Special Design. — Details. 

the line drawn from 6' to 7, as indicated by the arrows 
and draw a curve through the three points as shown. 
As the short fine 1 in the side elevation shows the con- 
necting line of the lip of the fender rim, this cut, as well 
as the established curve from to a, is developed to the 
pattern S as follows: Using 1 in the side elevation as 
radius and 1 in the pattern S as center, describe the arc 0, 
which intersect by an arc struck from a as center and 
a 0° in the true lengths as radius. The curve a in 
the side elevation is transferred to the pattern by taking 
the depth of the curve between the arrow points and 
placing it as indicated by similar arrow points in the 



464 



The New Metal Worker Pattern Book. 



pattern S, measuring from the center of the line a. 
Trace a line through points thus obtained in S, which will 
be the net pattern to which allowance must be made for 
seaming, wiring or bolting; below the line a 5' allow the 
necessary flange for turning down over the frame hue as 
desired. The pattern for the fender top is simply a 
rectangular piece whose girth is equal to. the distance 
from 1 to 7 in the side elevation, 10-inches wide, with a 










! J> 

1 1 
1 1 
1 1 

il 

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i i 
! 1 


1 
J 

1 

1 
1 

.1 
1 

1 
1 
I 
I 

1 

i 






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I 
1 
1 
1 
1 
1 
1 
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1 
1 
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1 2 


J 


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b 



Fig. 799. — Showing Development of Pattern. 

semi-circular end, as shown in diagram T. A section of 
the raised bead or swedge is shown in the pattern. Edges 
must also be added to this pattern for seaming or riveting. 
The pattern for the lip of the fender rim shown by X 1 
in the side elevation has been transferred to diagram, 
Fig. 799, as shown by 14 4' a in elevation. While this 
lip can be developed by triangulation, a simpler method 
can be used, that of parallel lines. Knowing that the 
true section on the line 4 1 is a true semi-circle 10 inches 
in diameter, place this profile in its proper position as 



shown which divide in equal parts, as shown by the small 
figures 1 to 4, on both sides. Through these intersections 
at right angle to 4 1 in elevation, draw fines intersecting 
4 1 in elevation at 1, 2, 3, and 4. From these intersec- 
tions parallel to 4 4' draw lines indefinitely, as shown, 
cutting the lower line of the lip at 1', 2', 3', and 4'. At 
pleasure at right angles to the fines just drawn, draw the 
line 4 b'. In similar manner draw the perpendicular 4 b 
in the semi-circle. Now, measuring from the 
fine 4 b in the true profile or semi-circle, take 
the various distances to points 1, 2, and 3, and 
place them on either side of the line 4 b' on 
similar numbered lines, thus obtaining the points 
of intersections 1 to 4 in P. Trace a line 
through points of intersections thus obtained 
which will be the true section at right to 4 4' 
in elevation. Using this section from which to , 
obtain the girth, the pattern is developed by 
drawing the girth line c d, parallel to 4 b', upon 
which place the girth of the section P. From 
these small figures perpendicular to c.d erect 
lines, which intersect by fines drawn from simi- 
lar points of intersections on the upper fine of 
the lip 1 4 in elevation, and parallel to c d, and 
through these points of intersections trace the 
miter cut 1° 1°. In a similar manner from the 
points of intersections 1' to 4' on the lower fine 
of the hp in elevation, parallel to c d, draw lines (not 
here shown), and obtain the lower miter cut, 1" 1". 
Now, using 1 a in elevation as radius and 1° on both 
ends of the pattern as center, describe the arcs indicated 
by a', which intersect in turn by arcs struck from 1" as 
Centers and 1' a in elevation as radius. This cut 1° a' in 
the pattern for lip will join 1 in the pattern for the 
apron, S, also 1 in the pattern for the l^-inch rim, 
shown in the side elevation. The rim pattern can be 
pricked direct from the side elevation. 



aH 



■J&* 



PROBLEM 237. 
Patterns for Mason's Chute. 



Numerous queries anent pattern cutting come to 
the journals aforementioned and in many of these queries 
tins book is mentioned, often by the remarks that the 
solution desired cannot be found herein. The solution 
presented here to one of these queries, evidently moved 
the author to certain reflections apropos the mentioning 



of this book and as these reflections are decidedly per- 
tinent and of value to readers of this book they are incor- 
porated in this demonstration, as was done in other prob- 
lems contained in this book. 

As has often been stated, it is to be borne in mind 
that this book is a collection of representative problems, 



Pattern Problems. 



If,-, 



some of which will, or rather its principles, apply to any 
new problem that may come up. However, it is a difficult 
matter for the average mechanic to apply the principles 
expounded for a similar problem to his particular case. 
This, of course, is no invidious reflection on the abilfty 
of the mechanic, because pattern drafting is a decidedly 
abstract science based on the higher mathematics, hence 
the innumerable problems that have appeared since the 
publishing of the last edition of this book. These prob- 
lems were not added to the book but are available in a 
series entitled "Practical Sheet Metal Work and Demon- 
strated Patterns" and the reader will find that a set of 
these books will materially help him in solving the prob- 



The axial line C S being continued indefinitely, the pro- 
file of the chute is built about it, as it were. Divide the 
curved part of the profile and project the lines from these 
division points to the miter line, as shown, finishing 
thereby that much of the elevation. Parallel to P C 
and at a distance of one-half the width of the hopper on 
the top fine of it, draw a short dotted line, as indicated 
by the clotted line at A 2 . With the trammels or com- 
pass set to 17 inches (the slant bight of the hopper), 
and with point 4° on the miter line as a center, describe 
an arc intersecting the dotted line, establishing thereby 
the point A 2 . From A 2 draw a horizontal line the width 
of the top line of the hopper as A 2 P A. Connect A 2 C, 




vation 



Plan 
Fig. 800.— Sketch of the Problem. 





Fig. S02. — Diagram of Triangles. 




Fig. SOI. — -First Steps for Obtaining Patterns. 



Fig. 803.— One-Half Pattern 
of Hopper. 



lems arising daily in his work. As his specific problem 
does not appear in any of the books aforementioned, and 
as it would be more difficult to show how to apply the 
principles of similar problems to the case, a demonstra- 
tion of Iris particular case appears herewith. 

The deign of the object as here given conforms as 
nearly as possible to his requirements as given in the 
sketch, specifications and other information furnished. 
To give the reader an idea of the design of the article 
various views are given in Fig. 800. 

As in Fig. 801, one draws a full-size side elevation of 
the object in this manner: Draw the center or axial fine 
P C S, P C vertical and C S at the required angle. Bisect 
the angle PCS, as shown, which gives the miter hne 
between the chute, or trough, and the hopper or bin. 



A C, and A B. Also A 2 to 3° and 2° with dotted lines, 
all as shown. No further work need be done on the ele- 
vation, so that now the developing of the patterns is in' 
order, and as follows: 

Therefore, for the chute pattern, continue the line 
R S T and place thereon the stretchout of the profile 
of the chute, that is to 4 and repeat, as shown. Draw 
the usual parallel fines which are intersected by the 
projectors from the miter dine. Tracing a line through 
the points of intersection completes the net pattern of 
the chute. 

Before the pattern for the hopper can be obtained 
it is necessary to ascertain the true lengths of lines 4° A 2 , 
etc., of the elevation and to accomplish this a customary 
diagram of triangles is generated as in Fig. 802. That 



466 



The New Metal Worker Pattern Book. 



is to say, on any horizontal line, the distances of the eleva- 
tion Fig. 801, as A 2 4°, A 2 3°, A- 2° and A 2 C are placed. 
For instance, keeping C of Fig. 802 constant for all spaces, 
the distance A 2 4° of Fig. 801 is C 4° of Fig. 802, and so on. 
From C, in Fig. 802, erect a vertical line equal in length 
to one-half the top line of the chute, or A 2 P of Fig. 801. 
It is to be remembered that tins space A 2 P can be taken 
because all four sides of the hopper are equal in width 
at the top ; in other words, the hopper is 20 inches square 
at its top. 

In Fig. 802 from 3° A and 2° erect verticals, making 
3° 3 X the same distance as 3 3' of Fig. 801 in the profile 
of the chute. And 2° 2 X in Fig. 802 like 2 2' in Fig. 801. 
2° A 2 of Fig. 801 like 0' of Fig. 801. Draw lines to X of 
Fig. 802, as shown, which completes the diagram of triangles. 

The first step in developing the pattern of the hopper 
is to draw any fine as B A in Fig. 803 equal in length to 
B A of Fig. 801. At right angles to B A in Fig. 803 
draw line B B° of a length coincident with 0' of Fig. 
801. Again, at right angles to B A in Fig. 803 and from 
A draw fine A A° equal to say A 2 P of Fig. 801 (one-half 
the width of the top line of the hopper). Connect B° 
and A° of Fig. 803. Set the compass to span A 2 X of 
Fig. 802 and with A° in Fig. 803, as center describe a small 
arc, which is in turn intersected by an arc described 
with B° as center and a radius equal to C B of Fig. 801. 
Connect lines as shown. From C of Fig. 803 as a center 



describe an arc of a radius equal to A° C of Fig. 803. 
From A° of Fig. 803 as center intersect this arc with one 
of a radius equal to the full width of the top line of the 
hopper, or say A 2 P A of Fig. 801. Connect A° A 2 
of Fig. 803. 

Employing A 2 of Fig. 803 as center, describe arcs of 
radii of the spaces 2 X X, 3 1 X, and 4° X of Fig. 802. 
Then from C of Fig. 803 as center describe an arc of a 
radius equal to the space C x H of the pattern of the chute 
in Fig. 801, the point of intersection in Fig. 803 being 
labelled H. Again, from this point H in Fig. 803 describe 
an arc of a radius equal to space H K of the pattern in 
Fig. 801, calling that point of intersection in Fig. 803, 
K. Now, from K, Fig. 803, as a center describe an arc 
of a radius equal to space K M of the pattern in Fig. 801, 
that point of intersection in Fig. 803 being designated as 
M. Trace a line through H K and M in Fig. 803 and 
join these points with A 2 as shown. From A 2 in Fig. 803 
describe an arc of a radius equal to one-half the top fine 
of the hopper, or A 2 P of Fig. 801. From M in Fig. 803 
describe an arc of a radius equal to A 2 4° of Fig. 801, the 
point of intersection in Fig. 803 being called 4 X . Connect 
M 4 X and 4 1 A 2 of Fig. 803, angle M i 1 A- being posi- 
tively square if the work is correctly done. This, then 
(Fig. 803), is one-half the net pattern of hopper. The 
small circles on the lines in the patterns indicate fines 
to be bent. 



PROBLEM 238. 
Pattern Details for Double Offset Y Branch. 



Two methods of developing the patterns will be 
required. The main pipe pattern can be developed 
by means of parallel fines and the branch pattern by 
triangulation. The first step is to draw the plan and 
elevation showing their relative positions from which 
the various views and miter lines are obtained. 

Let A 1 in Fig. 804 represent the profile of the 15-inch 
main pipe. Divide this into any convenient number of 
equal spaces as shown from 1 to 5 to 1. From 1 draw 
the horizontal line 1 B indefinitely and from any point 
as 1 in the side elevation draw the line 1 C at the required 
angle of 45 degrees. At right angles to 1 C lay off the 
distance of 12 inches or the diameter of the branch, as 
shown by C D. From D, parallel to C 1, draw a line 
which intersect by a horizontal fine drawn from the 
point 5 in the profile A 1 , thus obtaining the point 5 in 



the side elevation. Draw a fine from 5 to 1 which will 
represent the miter line between the branch and main 
pipe. Intersect this fine by fines drawn parallel to B 1 
from the numbers 2 to 4 in the profile A 1 . 

At pleasure in plan draw the profile A, a duplicate 
of A 1 , and divide it into the same number of spaces as 
shown, being careful that if 1 and 5 are placed vertical 
in A 1 , they will be horizontal in A in plan, thus making a 
quarter turn. From the point 3 in the profile A draw 
horizontal lines, which intersect by a line dropped from 
3 in the miter line 1 5 in the side elevation, thus locating 
3 and 3 1 in plan. From 3 in plan, at the 45-degree angle, 
draw the fine 3 8 indefinitely, as shown. 

At right angles to 3 8 from any point, draw the diam- 
eter 8 12 equal to 12 inches, which bisect and obtain the 
point a. Using a as center, with a 8 as radius, draw a 



Pattern Problems. 



467 



circle which divide into equal spaces, as from 6 to 13. 
From 12 draw a line parallel to 8 3, meeting the line 
drawn through the center of the profile A at 12. Now, 
from the various intersections in the profile A, draw 
horizontal lines which intersect by lines dropped ver- 
tically from the various intersections in the miter line 
1 5 in the side elevation. A line traced through points 



lines erected from d and c in plan, at e and c" respect- 
ively, as shown in the diagram. 

From c" in elevation draw the horizontal line inter- 
secting the vertical line previously drawn at d". The 
vertical hight is equal to d" e, which is used in obtaining 
the oblique view, and is placed from d' to e in the oblique 
view. Draw a line from e to c' in the oblique view at 




Fig. 804. — Plan and Elevation for Obtaining Miter Lines and 
Pattern for Y Branches from Horizontal Main. 



True Lengths of Dotted 
Lines in Top 

I T 6' i'B 



Elevation of 
-ff Branch 

6' 



True Length of Dotted Lino 
n Bottom 



Iff If 9 



JMV5 7 




H3-5-Z I! ISS&7II 

True Lengths of Solid Lines 
in Top 



Plan of Top 
Of Branch 




2? True Section on Joint Line 
between two Branches 



4 II ' W 

True Lengths of Solid Lines 
in Bottom 



ii e 



Half Paltern for Horizontal Pipe 
SftoTkX 3 

Profile of Branch , 





Fig. 805. — Method of Obtaining True Lengths of Lines for Developing 
Pattern for Forty-five-degree Branch. 



Pattern Details for Double Offset Forty-five-degree Y Branch. 



thus obtained as from 1 to 3 to 5 to 3 1 will be the miter 
line between the two branches and main pipe. 

Before the plan view of the upper end of the branches 
can be obtained an oblique view of the branch on its 
center line must be obtained as follows: Through a in 
the profile of branch H, draw the center fine a b on which 
establish any points as d and c. Equal in length and 
parallel to c d draw the line c' d' and extend d d' indefi- 
nitely. Through the branch in elevation draw the 
center fine e c" indefinitely, which intersect by vertical 



right angles to which through the point e draw the line 
6' 10', as shown. With t on the line c' e extended, as 
center, draw the profile of the branch pipe, H 1 , being a 
duplicate of H in plan. Divide H 1 the same as H. Make 
a quarter turn in numbering the intersections as shown. 
From the small figures 6 to 13 in H 1 draw lines parallel 
to t c', cutting the line 6' 10' as shown by similar numbers. 
From these intersections 6' to 10' in the oblique view, 
project lines in the plan parallel to e d' and intersect 
same by lines drawn parallel to d c in plan from similar 



468 



The New Metal Worker Pattern Book. 



numbers in the profile H. A line traced through points 
thus obtained will give the plan view of the upper end of 
the branch, shown from 6' to 13'. 

Trace this elliptical figure on the lower branch shown 
by 8 X 12*, although this is not necessary in practice, as 
only one branch is required. This completes the plan 
view, as shown. 

As the branch will be developed by triangulation the 
view of the upper end of the branch in elevation will be 
necessary. It is projected as follows: Extend c" d" in 
elevation to the right, as shown, and on this line place a 
duplicate of the oblique view in plan as shown by the 
similar letters c' d' e and numbers 6' to 13'. Parallel to 
c' d' in the elevation, from the various intersections 6' to 
13' draw fines to the left indefinitely, which intersect by 
fines erected from similar numbers 6' to 13' in plan, thus 
obtaining the points of intersections 6 to 13 in elevation 
as shown. Trace the elliptical figure shown which will 
complete the side elevation of the branch and main. 

A true section must now be found en the joint line 
1 12 in plan, which may best be found as follows: From 
the intersections 11 and 12 in the profile H in plan, draw 
lines parallel to the lines of the branch, until they inter- 
sect the joint line 1 12 in plan at 11 and 13. Now from 
the intersections 11, 12 and 13 in plan, erect perpendic- 
ular lines in the elevation winch intersect by lines drawn 
from points 11, 12 and 13 in elevation also parallel to 
the line of the branch, thus obtaining the points of niter- 
sections shown by 11°, 12° and 13°. Trace the profile 
through points thus obtained as shown from 1 to 13° to 
11° to 5, which will be the true section on 1 12 in eleva- 
tion. This completes the plan shown by F 3 8' 12' 12 12* 
8* 3* G and the elevation shown by B 1 6 11 5 E. 

For the pattern for the horizontal pipe, draw any 
girth line as K L at right angles to B 1 in elevation, on 
which place the girth of the profile A 1 , as shown by similar 
numbers. Through these small figures at right angles to 
L K draw the usual measuring fines, which intersect by 
lines drawn parallel to L K from similar numbered inter- 
sections on tfie miter line 1 5 in elevation. A line traced 
through points thus obtained as shown by 1° 1°° will be 
the miter 'cut and L 1° 5° 1°° K will be the pattern for 
the main horizontal pipe. 

To avoid a confusion of lines in obtaining the pat- 
tern for the branch pipe, a duplicate of the elevation of 
the branch shown by 1 6 11 5 with the various intersec- 
tions on same has been placed, as shown, by similar 
numbers in A in Fig. 805. In a similar manner a tracing 
of the top of the branch shown in plan Fig. 804, by 3 1 12 
12' 6' 8' has been placed as shown in B in Fig. 805, 



while a tracing of the bottom of the branch shown in plan, 
Fig. 804, by 3 5 12 12' 10' 8' has been placed in Fig. 805, 
as shown by similar numbers in diagram C. A tracing 
of the profile H of the branch pipe in plan Fig. 804 has 
also been placed in Fig. 805, as shown by similar num- 
bers in diagram H. 

The half pattern for the horizontal pipe shown by J 
with the various points of intersections on same, is a 
reproduction of a similar half pattern shown in Fig. 804. 
Thus the half plans B and C in Fig. 805 will show the base 
lines of the triangles winch will be constructed, with alti- 
tudes equal to the various hights in the elevation. With 
these true lengths which are to be found together with the 
true girth along 1 13° 5, in A, also in J and H, the pattern 
for the branch can be developed. The true lengths for 
the top of the branch will be found first as follows: 

Having placed the branch A at an angle of 45 degrees 
to the horizontal as required, draw a horizontal fine 
through all the intersecting points shown from 6' to 12' at 
the top, 1 to 5 at the bottom and through points 11°, 12° 
and 13° in the joint line or true section between the 
branches. At pleasure to the right and left of the branch 
draw the horizontal fines NO, PR, T S and M L. In 
the plan of the top of the branch B draw solid lines from 

2 to 7', 1 to 6' and 13 to 13' and connect by dotted fines 

3 to 7', 2 to 6', 6' to 13 and 12 to 13' or the shortest way 
to opposite points. Take the various lengths jaf the solid 
fines in B and place them on the fine N O, as shown by 
similar numbers. 

For example to find tne true length of 1 6' in plan B, 
take tliis distance and set it off on N O as shown from 1 
to 6', from which points erect perpendiculars until they 
intersect horizontal lines drawn from similar numbered 
points in the elevation A, thus obtaining the points 1 and 
6 in the true lengths. Draw a fine from 1 to 6 which is 
the true length of a similar numbered line in either plan 
B or elevation A. In this manner all the true lengths 
in diagram D are found. 

In a similar manner take the various lengths of the 
dotted lines in B and place them as shown by similar 
number on L M, from which points drop vertical fine 
cutting hnes drawn from similar numbers in the eleva- 
tion. Thus in diagram D are shown the true lengths 
of the solid lines and in diagram E, the true lengths of 
the dotted lines for the top of the branch. Connect the 
points by solid and dotted lines for the bottom of the 
branch as shown in diagram C. Take the lengths of 
the solid fines and place them on the line P R and the 
length of the dotted lines on the line S T. From the 
various numbers on these two lines, perpendiculars are 



Pattern Problems. 



469 



drawn, which will intersect similar numbered horizontal 
lines drawn from the branch A in elevation. Connect 
these points by solid and clotted lines. Then will the 
slant lines in F show the true lengths of the solid lines 
and the slant lines in G the true lengths of the dotted 
lines for the bottom of the branch. 

As the seam of the horizontal pipe was placed at the 
top in branch A, the seam should be placed along the top 
of the branch or on the line 1 6' in plan B. First, draw 
any line as 1° 6 in L equal to the true length 1 6 in D. 
With 1° 2° in the half pattern J as radius, and 1° in L as 
center, describe the arc 2°, which intersect by an arc 
struck from 6 as center and 6 2 in the true lengths in 
diagram E as radius. 

With radius equal to 6 7 in profile H and 6 in the 
pattern L as center, describe the arc 7, which intersect by 
an arc struck from 2° as center and 2 7 in the true length 
in D as radius. 

Proceed in this manner, using alternately, first the 
divisions in the pattern J, then the true length in dia- 



gram E, the proper division in the profile H, then the true 
length in diagram D until the line 5° 10' in the pattern L 
has been drawn. With radius equal to 5 11° in A, in 
elevation, and 5° in L as center describe the arc 11°, 
winch intersect by an arc struck from 10 as center and 10 
11° in the true lengths in diagram G as radius. With 
10 11 in the profile H as radius, and 10 in L as center, 
describe the arc 11, which intersect by an arc struck 
from 11° as center and 11° 11 in the true lengths in F as 
center. 

Proceed in this manner, using alternately, first the 
divisions in the true section on the joint fine between the 
two branches, shown in the elevation E, then the proper 
true length in diagram G; the proper division in the 
profile H, then the true length of the proper fine in dia- 
gram F, until the last fine I 1 Q 1 in the pattern L has 
been drawn, which is equal in length to 1° 6 in L or 
1 6 in D. Trace a line through points thus obtained 
in L. Then will 1° 5° l x 6 X 6 be the desired pattern. 
Laps must be allowed for seaming or riveting. 



PROBLEM 239. 

Designing Elbows for Rectangular Pipes. 



As has often been stated the first and most impor- 
tant essential in pattern cutting is to make a correct 
design of the object and to prepare as many correct views 
of the object that could possibly be of help to develop 
the patterns. Elbows are a leading article of sheet 
metal work and it would seem that there never can be 
too much written about them. Commenting not as a 
criticism but as additional discussion of some elbow 
problems herein, it is to be said a study of the elbow prob- 
lems appearing on pages 472 and 473 discovers some 
points to which attention is drawn. The elbows shown, 
especially the ones on page 472, which prescribe a curve 
for the back or heel, are ideal, inasmuch as a sweep 
is given for the shunting of the air from one fine 
of piping to the other. It could be said though that 
those designs give an excess of area, with a consequent 
waste of material, but this is not of importance as it only 
appears in the first cost of the outfit, while it might, 
however, cause friction. 

An experienced tinsmith will see that the parts must 
be joined by the common double seam along the corners 
of the elbows. An article so small as a furnace elbow may 
be rather difficult to make when it comes to turning the 



edges along the irregular outlines of the patterns. It is 
almost impossible to fit a stake head or dolly in the fitting 
to close down the double seam in the elbows shown 
on page 472, but perhaps the size of that elbow would 
permit the use of a stake of special shape, although ex- 
treme care would have to be taken in closing down the 
seam, owing to the irregularity of the corners. The 
popular Pittsburgh seam could not be employed on an 
elbow of that type, and it would appear that a more 
simple elbow could be designed for such a situation. 
Elbows similar to those of Figs. 816 and 817 must at 
times be constructed, irrespective of the form of the duct 
to which they connect. 

In at least one of the methods of laying out elbows, 
if an ordinary round pipe elbow is to be laid out, first a 
right angle is drawn, as shown in Fig. 806. Then, de- 
pending on the diameter, a quarter circle is described, 
using the apex of the angle or A for the center. If the 
elbow is large, as for blowpipe work, when it is made of 
as many as six to eight pieces, the throat and back will 
have a big sweep, and, by reason of having a large number 
of pieces, the sweeps of the throat and back will conform 
very nearly to a true circle, giving an ideal condition 



470 



The New Metal Worker Pattern Book. 



for a gradual turning of the direction of the air or other 
substance passing through it. 

In furnace work, however, a larger number of pieces 
than four is seldom used, and, for wall stack, the space 
available seldom permits more than three pieces. For 
that reason the quarter circle as in Fig. 806 is made as 
small as is practicable in putting it together. This 
quarter circle is divided into four equal spaces, as 1 to 5. 
From A through 2 a line is drawn which represents the 
rise of the miter hne. The remainder of the work of 
laying out elbows has been so often explained that it 
would be a waste of space to do it here, and this prelim- 
inary step is given because it is employed in developing 





Fig. 806. — Finding Rise 
of Miter Line. 



Fig. 807. — Procedure for Drawing 
Elevation. 




Side Elevatio. 



^^-^"1 


/ 






Plan 1 



Front Elevation 



Fig. SOS. — Complete Views of Elbow. 

the patterns of all elbows belonging in this class, which 
includes both plain elbows and transformers. 

If an elbow of the center reversible type, shown 
on page 273, Fig. 817, is to be laid out, so much of 
the preliminary as is shown in Fig. 806 is used. The 
next step is to draw the profiles in their correct positions, 
and complete the side elevation, as shown in Fig. 807, a 
plan and side and front elevation being given in Fig. 808. 
The back of the elbow is a straight line, and is easier, 
quicker and cheaper to make than the curved back that 
would be required by the design of Fig. 817, as shown as a 
dash hne in Fig. 807. Observe particularly that nowhere 
is the area of piece No. 2 reduced. It is understood that, 



if friction and its continual handicap is to be avoided, 
these elbows could be made of any number of pieces, 
which would make them very nearly similar to the ideal in 
the matter of friction-eliminating sweep. 

In joining the pieces of this elbow, there is a choice of 
seaming along the miter lines or the corners, but seaming 
along the corner admits of much more rapid and stronger 
work. However, the patterns would be essentially alike. 
In one case there would be two cheeks, with a throat 
in one piece, and a back, or heel, in one piece; whereas, 
in the other case, the pieces Nos. 1, 2 and 3 would be in 
one piece, with a seam at one corner, or wherever con- 
venient, and then joined at the miter lines. The pat- 
tern cutting differs but little in either case, so, assuming 


















A 


„3( 


G 


D 




B 


* ^~~~~"~-— — . 




Fig. 809.— Pattern Fig. 810.— Pattern Fig. 811.— True 

of Throat. of Heel. Length Diagram. 





e>^ 



Right Front Elevation 



Fig. 812— Pattern 
of Cheeks. 



Plan from left 
Fig. 818. — Right Reversible Elbow. 



that the seams are to be along the corners, a hne is drawn 
on which is placed the stretchout of the throat as shown 
by Fig. 809 and labeled E F G H. At right angles to 
this hne draw lines through these points. On E, and to 
both sides of the hne E to A, place one-half of the narrow 
side of the profiles, also do likewise with the wide side of 
the profiles at H. Finish the pattern, as shown, and 
follow similar procedure in Fig. 810 for the pattern of the 
back, or heel. 

For the pattern of the cheeks, the first step is to 
determine the true length of line G C, Fig. 807, which indi- 
cates the hne of bend of the cheek due to their having a 
warped surface at the piece numbered 3. Draw a hori- 
zontal hne equal in length to G C in Fig. 807, as G C, 



Pattern Problems. 



471 



Fig. 811. From these points erect verticals of a length 
equal to one-half the short side and long side of the pro- 
files, respectively. Connect these points with a line, 
which is then the true length of G C of Fig. 807. In the 
elevation of Fig. 807, the pieces No. 1 and No. 3, there 
shown, would also answer as the patterns of two sides 
of them, so anywhere redraw a duplicate of the elevation 
of piece No. 1, as shown in Fig. 812. From 2 in Fig. 812 
describe a short arc the diameter of which is equal to the 
length of Gl CI of Fig. 811. Intersect this arc by one 
having its center at B of Fig. 812 and its diameter of a 
length equal to B X, C X of Fig. 810, giving C as the 
point of intersection, in Fig. 812. From C in Fig. 812 
describe an arc of diameter equal in length to the miter 
line in Fig. 807 of piece No. 3. Intersect this arc by one 
having its center at 2 of Fig. 812 and its diameter equal in 
length to F X, G X of Fig. 809, making the point of inter- 




Side Elevation of both 




Front Elevation of 
Right Reversible El bow 



Plan of Left 
Reversible Elbow 



Fig. 811).. — Views of Right and Left Reversible Elbows. 



section at 3 in Fig. 812. Attach to line C 3 of Fig. 812 
the other part of piece No. 3 of Fig. 807, which completes 
the pattern for the cheeks. 

These are net patterns only, and to these the usual 
edges and laps must be allowed for seaming, etc. It is a 
good scheme to mark on the pattern where these different 
parts fit together, also where and how much the slight 
bends are to be made in the parts, as shown by the lines 
on the figures which have the cross-hatching, which means 
the inside. The rise of the miter line, as shown by Figs. 
806 and 807, is for any kind of a three-piece elbow, and 
hence the pattern for piece No. 1 will do for a three-piece 
elbow with the miter cut along the long side of the profile. 
Two pieces of No. 1 make a 45-degree offset with the miter 
along the wide side, and, similarly, with piece No. 3, 
elbows and offsets can be made having miters cut along 
the narrow side of the profile. Thus, a number of sets of 
patterns are obtained from one design or elevation of an 
■elbow, with the consequent saving of time in developing 



the patterns, which is of more importance with elbows of 
elliptical profiles. 

In Fig. 813 is shown a design for what is termed a 
" right reversible elbow." The side elevation for this 
design is obtained in precisely the same manner as for 
that of Figs. 807 and 808. The procedure for developing 
the patterns is essentially the same as for the elbow of 
Fig. 808, and, by consulting Figs. 807 to 812, the methods 
should be apparent. Note, however, that the side ele- 
vation of Fig. 813 is also the pattern of one of the cheeks 
designated A in the front elevation, omitting the warp 
line, inasmuch as the cheek is a true, flat plane and that 
a pattern for cheek B, only, must be developed. 

In Fig. 814 is shown a right, or a left, reversible elbow 
that also offsets somewhat to the side. The entire pro- 
cedure is identically the same as for the foregoing, and it 



Side Elev ation 





Front Elevahon.offsetting to right 



Ran, offsetting to left 

Fig. 815. — Combination of Ninety-degree Elbows. 

would seem unnecessary to present the drawings of the 
patterns. It is to be said, however, that the design, as 
here shown, is applicable only when the offset is slight, 
for the offsetting to the side squeezes in, as it were, the 
contour of piece No. 3, which means that the area is 
reduced. Projecting an oblique elevation may produce 
the means of obtaining miter lines that would avoid this 
restriction. However, it will be found that that will 
help but little by reason of the positions of the profiles, 
and that a complicated process is evolved which avails 
nothing in the end, for simple and rapid methods should 
always be followed. It is within the range of possibility 
that, in all cases, the offset could be produced some- 
where else in the line of piping than at the elbow, or else 
the offset could be made enough more to allow a com- 
bination of 90-degree elbows, as shown in Fig. 815. 
The patterns for these elbows are from pieces Nos. 1 
and 3 of Fig. 807, and these elbows can be made by either 
method explained before. 



472 



The New Metal Worker Pattern Book. 



PROBLEM 240. 
A Reversible Elbow Problem in Furnace Fittings. 



In furnace work it is often necessary to heat a room 
on the second floor by a wall register placed in the par- 
tition. The partition on the first floor, where the wall 
pipe runs, is not directly underneath the partition on the 
second floor, and is also at a right angle to the partition 
on the second floor. With the use of a flat elbow and a 
, reversible elbow these conditions can easily be met. 
Reversible elbows are usually made in three styles, center 
reversible, left reversible and right reversible, as shown in 
Fig. 816. In all three the methods of laying out the patterns 



two bends could be found by drawing a rear elevation, 
but the tinsmith will readily see the required angle by 
laying Ms parts together, leaving these two breaks until 
the last. Next, directly below the side elevation at Fig. 
818, draw a plan, as shown in Fig. 819, which is a view of 
the elbow looking down from the top. Then, draw a 
front elevation as shown in Fig. 820 as it would appear 
if looking at the front of Fig. 818. 

Develop the pattern for the front piece a b c d, as 
shown by Fig. 821, making the distance T the same as 



Fig. SIS. — Side Elevation of Center 
Reversible Elbow. 





Fig. 816.— Right Reversible Elbow. Fig. Si 7. —Center Reversible Elbow. Fig. 819. — Plan, 

s 



Fig. 822.— Pattern for Bach. 




3£, 
Fig. 820. — Front Elevation. 




I 1' 

i-z it 1 1 in Q 8 



Fig. 821. — Pattern for Front. Fig. 823 .—Pattern for Sides. Fig. 824.— True Length Diagram. 



are similar. Here is shown a method for drafting the pat- 
tern for a center reversible elbow 3| X 10 inch, as in Fig. 817. 
First, draw an elevation as in Fig. 818, making a 1 
the required width, say 10 inches and a b and 1 2 the 
required depth for collar flange, say 2 inches. Let b c 
be the desired size, say 4 inches, and let a b and 1 2 
(the required size for presenting the width of the pipe), 
be any desired size, say 3| inches. Then draw the bottom 
or curved back of the elbow to any curve desired. Extend 
ol c through until it meets the curved back and from this 
point draw a line b, then from b to 2, parallel to a 1. 
These lines represent the portion of the material which is 
bent to get the spreading effect of the sides, the angle of 
the bend for c 4 being taken from the front elevation, 
Fig. 819, at the point s. The exact angle for the other 



s 0, Fig. 820. Next develop the pattern for the back by 
first dividing off the curved back of the side elevation in 
Fig. 818 into equal spaces having one of the points come 
where the line extended from c intersects the back, also 
number the straight parts as 1 2 and 12 13. Now lay 
out a stretchout line N M, Fig. 822, having it come- 
directly from the center of the plan, Fig. 819, and lay off 
on this the spaces marked off on the back of the side ele- 
vation. Then drop vertical lines from the side elevation, 
Fig. 818, to the outside lines of the plan, Fig. 819, and 
carry them horizontally to vertical fines with corre- 
sponding numbers drawn from the stretchout Fig. 822, 
as shown by the lines carried from the points 8. Trace a 
line through these points of intersection. This line will 
be the pattern for the back. 



Pattern Problems. 



473 



To get the pattern for the side, first transfer a b and 
1 2 from Fig. 818 to one side, as Fig. 823. Then set the 
compasses to H L, Fig. 822, to get the true length of 2 4, 
Fig. 818. With 2 of Fig. 823 as a center, strike an arc. 
Then, with the compasses set to the true length of the 
line, b 4 of Fig. 818, and with b of Fig. 823 as a center, 
strike an arc cutting the first arc. Now, to get the true 
length of this line lay off a diagram of triangle as Fig. 824, 
in which b 4 is the same as 6 4 in Fig. 818 and X 4 the 
same as X 4 of the plan Fig. 819, b X then being the 
required length. Now, with the compasses set to s o 
of Fig. 820, strike an arc from b, Fig. 823 as center. 
Then with the compasses set to 12 4 of the plan, Fig. 819, 



and with point 4, Fig. 823, as center, swing an arc cutting 
the one swung from b. Draw a line from point 4 through 
c and on to d, making c d the same as c d of Fig. 818. 
Make d 13 the same as d 13 of Fig. 818, and 13 12 the 
same as Fig. 818. Now transfer each of the spaces from 
12 to 4 of the plan, Fig. 819 to the line c 4 of Fig. 823. 
Then drop lines at right angles from each of these 
numbers and make those lines the same length as the 
dotted lines on the side elevation, Fig. 818. Then draw 
the curved line from 2 to 12, Fig. 823, so it touches the 
outside points. The pattern is now complete for the 
side. Allow for flanges or folds, bend each side the 
opposite way when putting the fitting together. 



PROBLEM 241. 

Patterns for Sheet Metal Transforming Elbow. 



In the front and side elevation in Fig. 825 is shown a 
reproduction of the elbow and referring to the drawing it 
will be seen that the opening of the elbow at the bottom 
lies on a horizontal plane, the size of which is indicated 
by D E F G, and the opening of the elbow at the top hes 
on a vertical plane, the size of which is shown by 2, 2, 
14, 14 in the front elevation, the elbow forming an angle 
of 90 degrees. 

In this connection it is proper to say that it makes 
no difference what size the elbow may have at either 
end or at what degree it is set, the principles hereinafter 
set forth are applicable to any size or shape. In develop- 
ing the patterns the top and back, also the bottom and 
front, will be laid out by means of parallel lines, while the 
sides will be developed by triangulation. First draw 
the side elevation of the elbow, making 1 14 the hight of the 
upper pipe and 8 9 the width of the lower pipe, the 
quadrants 2 to 6 and 10 to 13 being struck from the cen- 
ters a and b respectively. Below the line 8 9 in side view, 
draw the full profile of the pipe, as shown by D E F G, 
which, however, can be omitted in practical work. 

To the left of the side view draw the center line A B 
and on either side of same proceed to draw the front 
elevation. In practice it will only be necessary to draw 
but one-half elevation. Make the distance from 8 to 8 
in the front elevation equal to E F of the profile. Project 
the corners 1 and 14 in the side elevation to the front, 
making the distance from 2 to 2 equal to the width of the 
upper pipe, and complete the profile of the top of the 
pipe, shown by 2, 2, 14, 14. From point 10 in the side 



view, where the quadrant runs tangent to the vertical 
line, project a line to the front as shown, and intersect 
this line at 10 and 10 by vertical lines erected from the 
points 8 and 8. Now, from 10 in the front elevation, 
draw lines to 13 and 2, as shown on either side of the cen- 
ter line. These lines represent the miter or joint hues 
forming the transition from a given size rectangular pipe 
at the bottom whose opening is placed horizontally to 
given size rectangular pipe at the top whose opening is 
placed vertically. The side and front elevations being 
correctly drawn the patterns are now in order. 

Divide the quadrant on the heel also the quadrant 
in the throat in equal spaces as shown by the small 
figures 2 to 6 and 10 to 13 respectively. Number the 
brake line 7 and the ends 1, 8, 9 and 14, as shown. Now, 
take the girth of the heel of the elbow from 1 to 8 and 
place it on the center line A B, as shown by similar num- 
bers, through which draw horizontal lines indefinitely, 
which intersect by lines drawn parallel to A B from the 
various intersections 1 to 9 in the front elevation, which 
points were obtained by projecting horizontal fines from 
similar numbered points in the side elevation. 

Trace a line through the various points of inter- 
sections thus obtained in the pattern, as shown on the 
left side from 1' to 8', and complete the full pattern shape 
by transferring the left hah pattern opposite the center 
line A B, as shown. Now, from the various points of 
intersection 9 to 14 in the side elevation draw horizontal 
lines to the left, cutting the miter line in the front elevation 
from 9 to 14, as shown. Take the girth of the throat of 



474 



The New Metal Worker Pattern Book. 



the elbow from 9 to 14 in the side and place it on the cen- 
ter lines A B, as shown by similar numbers. 

Through these small figures at right angles to A B 
draw horizontal lines, which intersect by lines drawn 
parallel to A B from similar numbered intersections in 
the front elevation. Trace a line through points thus 
obtained, as shown by 9' to 14' and trace the left half 
opposite the center line A B for the complete pattern. 
As the pattern for the side must be developed by triangu- 



Pattern for 
Sides of Elbow , 




Patte-rn for 
Bottom and Front-/ 
B ^ /0 



2 12 II 10 

True Lengths of 
Lines in 2-10- 13 
In Side Elevation 



True Lengths of 
Lines in 10-2-6 
in Side Elevation 



Fig. 825. — Patterns for Transforming Elbow. 

lation, a series of lines representing the bases of triangles 
must first be drawn in the side elevation, as follows : 

As the lines 7 10, 10 2 and 2 13, represent brake lines 
or lines on which slight bends must be made in the bending 
brake, then it will be best to draw the base lines of the 
triangles from 11 and 12 to 2; and from 3, 4, 5, and 6 to 10. 
The apex 2 in the side elevation lies on a plane indicated 
by 2 h in the front, and the various points 12, 11, and 10, 
in the side elevation rise above this plane, 2 h, in front 
elevation as much as is indicated by the points 12, 11, and 



10, on the miter fine 14 10 in the front elevation, and 
represent the altitudes of the triangles, which will be 
constructed as follows: 

In diagram V draw any horizontal line as c d, upon 
which place the length 2 10, 2 11, and 2 12, in the side 
elevation, as shown by similar numbers in V. At right 
angles to cd from points 10, 11, and 12, erect fines, 
making 10 10', 11 11', and 12 12' equal, respectively, to 
the distances measured from the line 2 h in front eleva- 
tion to points 10, 11, and 12. 

Lines drawn in V, from 10', 11', and 12' to 2 will give 
the true lengths of similar numbered lines in the side 
elevation. In a similar manner the apex 10 in the side 
elevation lies on a plane indicated by 10 i in the front 
elevation, and the various points 3, 4, 5, and 6, in the 
side elevation rise above this plane 10 i as much as is 




Fig. 826. — Perspective View of Finished Elbow. 

indicated by the points 3, 4, 5, and 6 on the miter line 
2 10 in front elevation, and represent the altitudes of the 
triangles shown in diagram W, which are obtained by 
drawing any line as ef, upon which place the lengths of 
10 3, 10 4, 10 5, and 10 6, in the side elevation as shown 
by similar numbers in W. 

At right angles to e / from points 3, 4, 5, and 6, erect 
lines, making 3 3', 4 4', 5 5', and 6 6' equal respectively 
to the distances measured from the line 10 i in the front 
elevation, to points 3, 4, 5, and 6. Lines drawn in W 
from 3', 4', 5', and 6' to 10, will give the true lengths of 
similar numbered fines in the side elevation. Having 
found the true lengths of the triangles, and as the divisions 
in the miter cut in the pattern for top and back are used 
for describing the edge line for the heel, in the side pat- 
tern, and the divisions in the miter cut in the pattern 
for bottom and front are used for describing the edge line 
for the throat, the pattern can be laid out as follows: 



Pattern Problems. 



475 



Take a reproduction of 7, 8, 9, 10 in the side elevation 
and place it as shown by 7', 8', 9', and 10' in the pattern 
for sides marked X. Now with 7' G' in the miter cut in 
pattern for top as radius and 7' in pattern X as center, 
describe the arc 6', which intersect by an arc struck with 
10' as center and 10 6' in diagram W as radius. Using 
radii 10 5', 10 4', and 10 3', in diagram W and 10' in pat- 
tern X as center, describe the short arcs 5', 4', and 3'. 

Now, set the dividers, equal to the spaces between 6' 
and 3' in the miter cut in pattern for top, and starting 
from 6' in pattern X step to arc 5', 4', and 3'. Again 
using the division 3' 2' in the pattern for top and 3' in 
pattern X as center, describe the arc 2', which intersect 
by an arc struck from 10' as center and 10' 2 in diagram 
V, as radius. Now, with radii equal to 2 11' and 2 12', 
and using 2' in pattern X as center, draw the small arcs 
11' and 12'. 

Set the dividers equal to the spaces between 10' and 
12' in the miter cut of pattern for bottom, and starting 
from 10' in pattern X step to arc 11' and 12'. Now, with 



the division 12' 13', in the pattern for bottom, as radius, 
and 12' in the pattern X as center, describe the arc 13', 
which intersect by an arc struck from 2' as center and 2 13 
in the side elevation as radius. The line 2 13 in the side 
elevation shows its true length, because it lies on a ver- 
tical plane, shown in the front elevation by 2 13. Now 
take a tracing of 2 1 14 13 in the side elevation, and 
transfer it to the pattern X, placing the line 2 13 in the 
side elevation upon the line 2' 13' in pattern X. 

Trace a line through points thus obtained as shown, 
then will 1' 8' 9' 14' be the pattern for the sides of the 
elbow. If the elbow is of such size that it will require 
heavy metal for its construction, then allowance must be 
made for laps for riveting purposes. If the elbow can be 
made of light material, edges must be allowed for seaming. 
In Fig. 826 is shown a perspective view of the elbow, laid 
on its back, to more clearly show the miter joints. The 
numbers shown from 1 to 14 are similar to those shown in 
the patterns, while B. L. denote where the brake lines 
take place. 



PROBLEM 242. 
Ornamental Roof Flange Pattern for Flag Pole. 



In discussing the procedure for obtaining the pat- 
tern for a flag pole on an ornamental roof, it might be 
said that the problem appears to be an easy one, but all 
the trouble is in the development of the patterns for the 
sides. The top is to be 4| inches square. The profile 
for the front is indicated by A and for the back by B. 
A reduced reproduction of the side elevation of the 
problem is shown, in Fig. 827, above the plan in the accom- 
panying illustration in which a side view of the desired 
molded flange is given. 

In this problem the patterns for the front and back 
will be developed by parallel lines, while the two sides 
must be developed by triangulation. After the side 
elevation is drawn, as shown, a plan of the flange must be 
drawn directly in line with the side elevation, the top 
opening to receive the flag pole being 4i inches square. 
The first step is to obtain the patterns for the front A 
and back B, but as the part above arrows a and b is simply 
a square miter cut, this portion will be omitted in the 
solution. In practice the pattern for the upper head can 
be joined to the front, back and sides, before cutting and 
forming up. 

Divide the given profiles A and B into any desired 



number of spaces, being careful to make the spaces closer 
where the mold is smaller, all as indicated from 1 to 6 in 
A and 7 to 12 in B. At right angles to the plan of the 
front draw any line as C D, on which place the girth of 
the profile A in elevation, as show by similar numbers 
1° to 6°. Through these points and at right angles to 
C D draw the usual measuring lines, which intersect by 
lines drawn parallel to C D from similar nmnbered inter- 
sections on the miter line in plan, which were previously 
obtained by dropping lines from similar numbers in the 
profile A in elevation. Trace a line through points thus 
obtained, as shown from 1 to 6 and transfer this half 
pattern opposite the has C D. The various divisions 
along the miter cut 1 to 6 will be used in obtaining the 
pattern for the side. 

In a similar manner from the various intersections 
7 to 12 in the profile B in elevation drop lines, intersecting 
the miter line in plan, also indicated by similar numbers. 
Proceed as was done in obtaining pattern A for front A 
until the full pattern B is obtained as shown. The divi- 
sion along the miter cut, 7 to 12, will also be used in devel- 
oping the side pattern. 

As the pattern for the side must be developed by 



476 



The New Metal Worker Pattern Bock. 



triangulation, both in plan and elevation, connect the 
opposite points by lines, as shown by the solid lines 
drawn from 4 to 9, 3 to 10 and 2 to 11; and by dotted 
lines from 5 to 9, 4 to 10, 3 to 11, and 2 to 12. Then will 
these solid and dotted lines in plan represent the bases of 
sections which will be constructed with altitudes equal to 
the various vertical nights shown by similar numbers in 
the elevation. 

To obtain the true lengths of the solid lines shown in 



True Profile through 
X-YinPlanorElevatra« 




Pattern for Front A 



Fig. 827. — Pattern for Ornamental Roof Flange for Flag Pole 



plan, proceed as follows: From point 5 in elevation draw 
the horizontal line 5 H. As 5 8 and 1 12 in elevation show 
their true lengths, no sections need be developed for these 
lines. Now, take the lengths of the solid lines in plan as 
4 9, 3 10 and 2 11, and place them on the line 5 H, as 
shown by 4' 9', 3' 10' and 2' 11', starting from the left and 
leaving space between 4' 3' and 2'. From these points of 
intersections erect perpendicular lines which intersect by 
lines drawn parallel to 5 H from similar numbered points 
in both profiles A and B, as shown by the points of inter- 



sections 4, 9, 3, 10, 2, and 11, in the true lengths. Solid 
fines connected from 4 to 9, 3 to 10, and 2 to 11 will show 
the true lengths of similar numbered fines in either plan 
or elevation. 

In a similar manner take the various lengths of the 
dotted fines in plan as 5 to 9, 4 to 10, 3 to 11, and 2 to 12, 
and place them on the horizontal fine 5 H, beginning at 
point 5' and leaving a small space between starting points, 
as shown by 5' to 9', 4' to 10', 3' to 11', and 2' to 12'. 
From these intersections erect vertical 
lines, intersecting them by horizontal 
fines drawn from similar numbers in 
both the profiles A and B, resulting in 
the points of intersections 5', 9, 4, 10, 3, 
11,2, and 12. Dotted fines drawn from 
5' to 9, 4 to 10, 3 to 11, and 2 to 12 
will show the true lengths of similar 
numbered lines in either plan or eleva- 
tion of the diagram. 

The pattern is now in order and is 
developed as follows: Take a tracing 
of 5 6 7 8 in the side elevation and 
place it as shown by 5 6 7 8 in the 
pattern for the side. With 8 as center 
and 8 9 in the miter cut in the pattern 
for back as radius, describe the arc 9 
in the pattern for side. With the true 
length 5' 9 as radius and 5 in the 
pattern for sides as center intersect the 
previous arc at 9. With radius equal 
to 5 4 in the miter cut in the pattern 
for front A and 5 in the pattern for sides 
as center, describe the arc 4, which in- 
tersect by an arc struck from 9 as cen- 
ter and the true length 9 4 in the 
diagram as radius. Proceed in this 
manner, using alternately first the divi- 
sions along the miter cut in the pattern 
for back B, then the proper true length 
in the diagram of true lengths; the 
proper division along the miter cut in the pattern for 
front A, then again the proper true length, until the line 
1 12 in the pattern for the sides is placed, which is made 
equal to 1 12 in the side elevation, its true length. A 
line traced through points thus obtained as shown by 
1 6 7 12 will be the desired side pattern. 

The square miter pattern for the bead above the 
arrows in the side elevation must be added to the patterns 
for the front, sides and back on the fines 1 1, 1 12, and 
12 12, respectively. When forming up the sides one 



Pa+ternforBackB 



Pattern Problems. 



477 



must be formed right and the other left. They must be 
bent or molded so that they will miter with the front and 
back pieces. If, however, a profile is desired so that the 
sides can be accurately formed, the method of finding the 
true profile through the center X Y in either plan or ele- 
vation is as follows: 

Where the solid and dotted lines in elevation cross 
the center line X Y as indicated by the heavy dots, take 
a tracing of these intersections, numbering each dot to 
correspond with the number of the line passing through 
X Y and place these numbers on the vertical line X 1 Y 1 , 
as shown in the true profile. In a similar manner where 



the solid and dotted lines in plan cross the center line 
X Y, also indicated by the heavy dots take these various 
distances with the number of the lines which they rep- 
resent and place them on the horizontal line X 2 Y 2 , as 
shown. From the various intersections on the vertical 
line X 1 Y 1 , draw horizontal lines, which intersect by 
vertical lines, erected from similar numbers on the hori- 
zontal line X 2 Y 2 , resulting in the points of intersections 
shown. A line traced through these points will be the 
true profile through X Y in either plan or elevation, and 
represents the profile to be used along the center of the 
pattern when forming up the sides of the flange. 



PROBLEM 243. 



Pattern for Irregular Shaped Box. 



Although there are hundreds of problems of a diver- 
sified nature worked out herein, the book could not pos- 
sibly contain all the problems that may arise owing to 
the great number of articles made from sheet metal. 
Oftentimes the book is misjudged by the statement that 
it does not give a rule for a special case inasmuch as 
there are really but three principles or rules in all pattern 
cutting, viz.: Parallel development, radial development 
and development by triangulation. All these are thor- 
oughly elucidated and some one or all would apply to 
the solution of a problem. 

Many artisans are not well versed in descriptive 
geometry and accordingly are not familiar with the fun- 
damental principles of the systems of surface develop- 
ments mentioned, consequently they cannot apply the 
rule or rules to their problems when guided by the explan- 
ations of similar problems in the book. This is no dis- 
credit to them as everyone cannot have the necessary 
preliminary training to be masters of pattern drafting 
and it is, therefore, imperative to expound their problems 
step by step. 

It is best not to refer to a like problem in the book for 
a hint to the procedure unless it can be said that the 
problem referred to is identical with that required and it is 
most desirable that all owners of the book make an 
effort to grasp the fundamentals of pattern cutting by 
diligent study of the first parts of the book for they will 
then more readily grasp the solution of the later problems. 

To solve this problem draw a rectangle A B C D to 
the required size as in Fig. 828, which represents the out- 
line of the base of the object in plan. Also draw another 



rectangle for the top as shown by E F H G. The side 
elevation is represented by the trapezoid labeled I J L K 
and the front by the one marked M N P O, which also 
outlines the back or rear elevation. By drawing the hip 
lines F B and H D the plan is completed. 




Fig. 828. — Method of Obtaining Patterns for Parts of Box. 

The object has four sides, the only one of which is 
shown true is the back. This means that the front eleva- 
tion will do for the pattern of the back. 

To show the front of the object true, that is to develop 
the pattern, extend the center fine a & of the plan indefi- 
nitely and place thereon the length J L of the side eleva- 
tion as indicated by c d. At right angles to the line a b 



478 



The New Metal Worker Pattern Book. 



and fromc and d draw lines equal to B D and F H, as 
shown by e f and g h. Connecting these points makes 
efgh the pattern of the front. 

For the sides draw a line equal in length to and parallel 
to C D as i j. Extend the line G C downward and from 
i place the distance M N (or OP), of the front elevation 
as shown by i k. From k and parallel to i j draw a line 
of the same length as G H which is k I. Draw a line 
from I to j, which completes the pattern of the sides. 

The angle of the back and side is square, while the 
angle of the side and front is unknown. To find it, 
make an oblique elevation of the hip F B by drawing any 
fine as m n parallel to it and of its length. At right 
angles to m n draw the line m o equal in length to I K 
of the side elevation. Connect o and n, which then is the 
oblique elevation of the hip. At right angles to F B 
draw a fine from the side to the front as 2 3, which project 
to the base fine of the oblique elevation and mark the 
point 4. At right angles to the line o n and from 4 draw 
the line 4 5. Through point 4 and parallel to o n draw a 
fine and make 4 6 to the length of 2 9 of the plan and 4 7 
to the length of 9 3. Then 6 5 7 will be the true angle of 
the side and front. 

To have the patterns of the different parts of the 
object in one piece with the seam through the center of 
the back, redraw the pattern of the front anywhere as 



e' g' In! f in Fig. 829. To the sides of this connect the 
patterns of the sides as h' f i' j' and e' g' k' V. To these 
connect one-half of the back as V k' n' w! and i' j' p' o'. 
This gives the full pattern of the object. 

Along the outer hne w! to o' allow the wire edge 




Fig. 829. — Pattern in One Piece 

required and also allow similar edges on the inner Hne n! 
to p' and on the fines o' p' and m! n'. 

In bending up the object square bends are made on 
the fines V k' and i' j' and bends to the angle 7 5 6 of 
Fig. 828 on lines e' g' and /' h'. 

In conclusion it may be said that the rule employed 
is a combination of the parallel and triangular systems, 
principally the parallel. 



PROBLEM 244. 



Pattern for Compound Curve Elbow. 



As a rule most all these problems emanate from readers 
of journals published by the publishers of this book, who, 
having a problem and unable to find a solution, appeal 
to the editors or otfier readers of these journals. The 
query pertaining to this case stated that his sketch Fig. 830 
has inlet A rectangular in shape, but makes a quarter 
turn to outlet B. Incidentally, the reader stated further 
that he was unable to find a solution of similar nature in 
this book. 

A study of the sketch leads to several inferences as 
to the requirements and it could be said that in heating 
and ventilating work having rectangular ducts, such a 
turn and offset would commonly be negotiated by the use 
of six ordinary elbows, two 90-degree elbows and four 45- 
degree offsets. It has been stated, however, that the 
elbow is to be of copper and from a close inspection of the 
sketch it would seem safer to presume that the object 



is for leader, that is to say, roof drainage work, or some 
such similar purpose. 

The sketch in plan, indicates that narrow side and 
narrow side of outlet and inlet and wide side and wide 
side, are not to connect, but just the reverse, wide side to 
narrow side. It would be more proper to twist narrow 
side to narrow side and wide side to wide side, like in Fig. 
831. Such an object then comes within the range of 
spiral problems in the nature of the spiral package chute, 
and a very clear description of the pattern developing for 
an elbow of this land can be found on page 32, Vol. 1, of 
" Practical Sheet Metal Work and Demonstrated Pat- 
terns." In recommending that demonstration the cau- 
tion, or rather reminder, accompanies it that the two sides 
which are developed by triangulation are distinctly 
warped surfaces and require a certain amount of stretch- 
ing or Humping, to bring them to their correct contour. 



Pattern Problems. 



479 



Another supposition that may be formed from the sketch, 
Fig. 830, is that the quarter turn and double offset could 
be made by utilizing straight sections of the pipe, as 
shown in the plan of Fig. 830. Although this supposition 
is refuted by the data furnished and the elevation of the 
sketch, a procedure for obtaining the patterns for such 
an elbow is explained herewith: 

In Fig. 832 a plan was drawn indicating thereon the 
outlet and inlet which have the proper amount of offset 
in two directions and the quarter turn, all as prescribed 
by the plan of Fig. 830. In the development of the pat- 
terns for tins problem an oblique elevation is necessary 
and to acquire that elevation it is customary to employ 
one line as a basis to work on. In a cylindrical object 




Fig. 830. — Curve in Elevation. 



the axial line is used, while for objects having, say, a 
rectangular profile or shape, an edge or corner line is used. 
Therefore, line 4° 4 in the plan of Fig. S32 is used as a 
basis to project the oblique view, as shown. Continue 
the line 4 C and make F C the hight of F' C in the 
front elevation. Lengths A B and C D being as con- 
venient, bisect angles ABC and BCD which will give 
the miter lines C G and B H. Project line ABC and D 
to front elevation and thence to the side elevation. 

In the plan connect points 1° 1, etc., and also oppo- 
site points as 1° 2. Project the plan view of the oblique 
view as shown and also to the front and side elevation, 
carrying the hights of the lines in oblique view as 1° I s , 
etc., on the dividers to the elevations as 1°° l 60 and l 0o ° 
1 6 °° and l c l" to l''r° and 1" l co ° in the elevations, etc. 



The patterns for the vertical pieces of the elbow are 
obtained by the parallel line system from the oblique 
elevation, as shown, at X. Before the pattern of the mid- 
dle or offsetting piece can be developed the correct length 
of the corner lines and the lines connecting opposite 
corners must be found. To the right of the elevations 
project lines indefinitely from all the points in the eleva- 
tions as l"° l co C B', etc. Erect two convenient vertical 
fines as shown in the diagram of triangles. On those 
lines and in diagrams marked P and marking always 
from the vertical line, set the distances in plan of the corner 
fine, 1 ° 1 in plan is l 1 1" in diagram P, etc. Connect the 




p^^^\^^ w 




Fig. 831. — Curve in Elevation and Plan. 



points thus located to proper points on the vertical line 
as 1™ to l xxx , etc. Do likewise on diagram R, the dis- 
tances, as l r and 2 TT , being taken from the fines in plan 
joining opposite corners, as 1° 2. 

For the pattern as shown at K, draw any fine as 1 ° 1 
of a length equal to l xx l xxx in diagram P. Then with 
the point of the compasses placed at 1 °, in K, and set to the 
distance 2 X 2 XX in R, strike a small arc. Then with the 
dividers spaced to 1 2 in the plan, intersect this arc giving 
point 2 in K. Repeat this procedure until the full pat- 
tern is obtained. 

It should be remembered that the surfaces of this 
object are what is termed " warped surfaces," hence 



480 



The New Metal Worker Pattern Book. 



slight bends should be made on R lines. However, the 
amount of warp in this object is so small that it really 



elevations. On page 39 of Vol. 1 of " Practical Sheet 
Metal Work and Demonstrated Patterns," will be found a 




Fig. 832. — Patterns for Three-Piece Sectional Elbow. 



could be ignored and only the square bends made on lines 
P. Laps should be allowed as required. 



1 
1 
1 

b, a 




1 
c 


r 









Fig. 83/,.— Plan Diagram of Fig. 833. 

As was said in the foregoing, the requirement in this 
case is for an elbow that must be curved when viewed in 



problem of like nature, except that it does not make the 
quarter turn. The plans as shown in that book are not 
correctly represented, strictly speaking, for when the 
development ignores the plan, so to speak, straight lines 
in plan are not obtained, but the result is an irregularly 
curved line or lines in plan. As an additional aid to the 
study of such a solution of the problem, refer to Fig. 833 
of the accompanying drawings. 

As was mentioned the plan is not required in this 
scheme of development, but to make this demonstration 
somewhat different than the others, the amount of offset 
one way is made greater than the other way. Supposing 
the amount of offset is, as shown in Fig. 834, by a & and 
b c, it is transferred to the horizontal line of the elevation 
in Fig. 833, as shown, and no further use is made of the 



Pattern Problems. 



481 



plan for only the profiles of the inlet and outlet are needed 
and are placed as shown. The hight of the elbow a M 
is as required and having established points A and M, a 




Fig. 833. — Using Curve in Elevation Only. 



graceful compound curve is sketched in free hand, as in 
this drawing, or by means of the compass as in Fig. 830. 
Set the dividers to a convenient space and starting at M 
step off spaces, letti: lg the final space B A be unequal if it 



across to the right. Establish points M' and A' in the 
front elevation and also M" and A", and M'" and A'" in 
the side elevation. Draw similar curves to these points 



-iW 




Plan 

Fig. 835. — Zolu'ion Conforming to Requirements. 

as shown, being careful to keep the distance apart of 
each side of the object, never less than 1 2 of the 
profiles. 

In doing this it is well to remember that it is best to 



so happens. Draw horizontal lines from these points I err in the matter of having greater area in the elbow than 



482 



The Neiu Metal Worker Pattern Book. 



to have a restricted area, and so it might be said of Fig. 
833 that it would have been better if the outlines of the 
front elevation, at about J, K, and L, were a trifle farther 
apart, to give greater area. 

The patterns for the four sides are obtained by the 
parallel line method. That is to say, stretchout A to M 
of the front elevation is set on line A to M above the side 
elevation and stretchout A' to M' (taking each space 
separately as they all differ), is set on the line A' to M' 
and the usual parallel lines are drawn as shown. These 
lines are intersected by lines projected from corresponding 
points on the curves in the side elevation so that pattern 
Y is for the side 1° 4° to 1 4 of profiles in the front eleva- 
tion, while Z is for 2° 3° to 2 3. Similarly, X is for 3° 4° 
to 3 4 and V is for 1° 2° to 1 2. Bear in mind also that 
stretchout A" to M" is above the front elevation as also 
is stretchout A'" to M'". 

In shaping these parts it is to be remembered that 
the stay (for instance, for V would be curve A'" M"' of the 
elevation), should always be held at right angles to 
the parallel lines as shown by the line g and h of V, for the 
idea is that V is simply cut out of a piece of metal like A'" 
k m and M'", which is curved to the shape of line M'" A'" 
of the side elevation. 

A solution of the problem conforming, so it would 
seem, the nearest to the requirements as stipulated, is 
presented in Fig. 835. This is nothing more than a manip- 



ulation of the principles of projection to obtain an object 
that is curved in both views of the elevation, although 
having straight lines when viewed in plan. 

The plan was drawn identical to Fig. 831 and the 
curve AM in the front elevation drawn at pleasure and 
divided into spaces as shown. These points were dropped 
to the plan on lines 1° 1 and 2° 2 carried horizontally 
■across to line 3° 3, vertically back to line 4° 4 and then 
horizontally back to line 1° 1, meeting the same points in 
that line. 

Lines were drawn horizontally across to the right in 
the elevations from the points A to M. These were inter- 
sected by lines projected up from the plan on line 3° 3, 
which gave the curve A' M'. Similarly, the points on 
lines in plan were projected to the right to line S T and 
swung around, using S as a center, to line S W, thence 
upward to intersect hke lines from curved line M A, 
giving the curved lines A" M" and A'" M'" or the side 
elevation. Patterns X, V, Y, and Z were obtained by the 
parallel line system just as was done for Fig. 833, this solu- 
tion differing from that one only in the fact that the plan 
was not ignored but employed to govern the elevation. In 
conclusion it might be stated that in this book a problem 
giving the principles that could be applied to the solution 
of this problem is No. 48. In that problem, however, the 
profiles are differently situated than in his case and would 
therefore require skill in adjusting to this case. 



PROBLEM 245. 



Offsetting Elbow Problem. 



The exigencies of the exacting requirements of the 
sheet metal trade give birth to many perplexing problems, 
one of which follows: 

It rises on the elevation 30 degrees and then turns 
away on the plan 40 degrees. The problem presents an 
interesting study in projection and developing surfaces 
of solids. There are probably many ways of solving 
problems like these, but no matter how solved it is a 
formidable task. 

Authorities differ on the many systems of projection, 
but without arguing as to which is better, the third angle 
or the first angle, and a lot of other technical premises, 
it can be said that the system best understood by sheet 
metal workers is that in which the general view is called 
an elevation and a view looking down on the object 



and placed below the elevation is called a plan, as pre- 
sented in Fig. 836. Perhaps this would be called first 
angle position, but it all is a little too deep for busy 
mechanics. 

In developing the pattern required, the first essential 
is to consider a certain imaginary line, known as an axis, 
about which the body proper of each piece of the object 
is built, because it is impossible to tell at once just what 
shape and position these pieces will have and this axis 
line is a central fine of the pieces, shown in these diagrams 
by the heavy solid line, and can be readily located. 

Presume that in plan of Fig. 837, the oval shown is 
as required and that the axis line A B, indicates how and 
how much the elbow turns away in plan. Projecting 
this fine to an elevation, that is to say, to a view on lines 



Pattern Problems. 



483 



of sight, the direction of which is shown by arrow pointer 
x, the axis line would be A' B' in elevation and prescribes 




Fig. 836.— Sketch of the Problem. 




Oblique Elevation 



Fig. 837. — Orthographic Study of the Proposition. 

the amount of rise in the elevation. The axis line of 
the oval piece in elevation, is labeled A' C and can be 



as long as desired. Neither the elevation nor the plan 
show the axis line A B in its true length because in both 
views one end of the line leans away from the observer 
and therefore shows foreshortened. To see this line in 
its true position and exact length it is necessary to pro- 
ject a view along arrow pointer X Y, called by most 
pattern draftsmen an oblique elevation. 

To do this, first draw a horizontal fine in the elevation, 
from point A' as shown by the heavy dotted fine, terminat- 
ing it where it touches the projector line from B in plan 
and calling it point D. At right angles to the axis line 
in plan and from points A and B, project lines indefinitely. 
Draw line A" D" parallel to line A B and then take the 
distance D to B' in the elevation place it from D' in the 
oblique view obtaining thus, point B". Connect A" to 
B" which will be the true view and length of the axis 
fine A B. 

Make C" A" m the oblique view equal to C A' in the 
elevation and then bisect the angle C A" B" in the oblique 
view, as directed by points b d and e. A line from e 
through A" bisects angle C A" B" and is the edge view 
of the cutting plane between the pieces of the elbow, or 
the miter fine. 

Continue line A" B" in the oblique view and place 
thereon the profile of the round end of the pipe as shown 
by 12 3 4. Divide this profile into equal spaces and 
draw the fine 1' 3' at right angles to fine A" B" and pro- 
ject to it the divisions in the circular profile as shown. 
Divide the profile in plan also, and project the division 
points up to the miter fine and connect points 69' and 1' 
and 91' and 3' also draw the line g h. 

Place a duplicate of the circular profile on the ex- 
tended line A B in the plan as shown by 
1°, 2°, 3° 4°. Project the division points 
of this profile into the plan to intersect like 
numbered projectors from line 1' 3' in the 
oblique view. Like this: 1' in the oblique 
view is to intersect at V" in the plan and 
so on, resulting in the elliptical view in 
plan of the circular end of the elbow. 
Complete the plan by drawing the two 
outside lines as shown. 
Project the division points upward to the elevation, 
then draw the horizontal fine through point C. Measuring 
always from this line and taking the various lengths of 
the same lines from fine g h to the miter line 69' 91' in the 
oblique elevation, obtain thus the elevation view of the 
oval piece of the elbow. For instance, point 100 in plan 
is fine k 100' in the oblique elevation. This length is 
placed in the elevation on line from point 100 in plan, as 



484 



The New Metal Worker Pattern Book. 



k' 100 J . The elliptical view of the circular end of the 
elbow is obtained in precisely the same manner, measuring 
from line D' in the oblique elevation and line D in the 



they are placed in the plan of Fig. 837 as 50 100 base 
and apex 3", also 70 80 base and apex 1" ' and projected 
to the elevation and oblique view as shown. 

All the foregoing was absolutely necessary to study 
the proposition and is a sort of preliminary survey. 
Knowing just what is wanted, a good deal of that work 
could be dispensed with; in fact, the many projector and 
working lines will make a terrible mix-up in the further 
developing of the solution. So then, as in Fig. 838, turn 




Fig. S3S.—A Duplicate in Part of Fig. 837. 

elevation which completes the different views of the object 

and ought to tell just how the object will actually appear. 

Inasmuch as there are flat parts to the oval profile, 

which by the way, should be termed a rectangular figure 



F*&l ?c* x 




6-r^EDCB'J 
n 

Fig. S39. — First Set of Solid Lines. 



with same circular ends, there will be flat triangular 
portions in the transformation pieces of the elbow. The 
apex of these triangles can be any one of the division 
points on line 1' 3' in the oblique view, it being immaterial 
owing to the transformation piece being decidedly twisted, 
no matter how it would be designed. Therefore, follow- 
ing one's judgment for the best position of these triangles 



the axis line A B completely around to facilitate things in 
general and in correct position, to duplicate Fig. 837, 
place the profile of the oval as shown. This is really all 



-""-- — — &£T 



GffFEDC 



Fig. 840.— First Set of Dotted Lines. 

that is needed of the plan. The original elevation of the 
axis fine is also shown in Fig. 838 and this much is all that 
would be drawn, supposing that Fig. 837 had not been 




Fig. 841. — Second Set of Solid Lines. 



10, 
11-9- 

IZ-8 



1-7 



.\M X 
,N* 






Fig. 842.— Second Set of Dotted Lines. 

developed. Only the axis lines and line 7' 1' also elevation 
of the oval piece and miter lines as shown, is required 
in duplicate of the oblique elevation of Fig. 837. 



Pattern Problems. 



485 



The first pattern to develop is that for the oval piece 
and the procedure is to extend line g h, Fig. 838, and place 
thereon the stretchout of the oval profile, these division 
points having been rechristened A to to prevent con- 
fusion with the symbols of the circular profile. From the 
points A to A of the stretchout erect vertical lines. 
Project lines across from the miter line to intersect like 
lines of the stretchout. For instance, K' on the miter 
line is projected across to intersect line from K° in the 
stretchout giving point K°°. A line drawn through these 
points completes the net pattern of the oval piece. 

The transformation piece will naturally have to be 
triangulated to develop its pattern. Now, as the trans- 
forming of the shape takes place in the entire girth of 
the object the entire pattern must be developed, whereas 
if the object was in symmetrical halves or quarters, one- 
half of the pattern would do for the other half or one- 
quarter for all quarters. Such being the case it requires 




6-F 



Fig. 843. — Reproduction of Part of the Oblique Elevation of Fig. S3S. 

additional patience and care because triangulation is the 
most tedious process imaginable. The precise definition 
of triangulation is, having the base fine and the altitude 
of a right-angled triangle, the hypotenuse gives the 
required true length of a line, shown foreshortened in the 
views. This method has been modified in several differ- 
ent systems and a system ideally suited for this problem 
is where the line in the oblique elevation is considered 
the distance and the base fine between two parallel fines, 
the length of one of these parallel fines is taken from one 
profile, while the length of the other is taken from the 
second profile, to wit: 

Point 4 1 in Fig. 838 is the same as point 4" in Fig. 
837, and is only used here to explain the system. Line 
E' to 4' in the oblique elevation is the base line spoken of 
and is placed where convenient, as from 1 7 to E of Fig. 
839. Vertical lines are erected from points 1 7 and E 
in Fig. 839 and are the parallel lines mentioned. The 
distance B 4 1 of Fig. 838 (or what is the same thing, 10 
to 10° of the circular profile) is placed from 1 7 to 4 of 



Fig. 839. Also, space E to E 1 of the plan, Fig. 838 is 
placed from E to E 1 in Fig. 839 so that the fine from 4 
to E x in Fig. 839 is the true length of line E' of the 
oblique elevation of Fig. 838. In other words, the figure 
bounded by the corners B E 1 E and 4* of Fig. 838 is placed 
in Fig. 839 as shown by 17EE 1 and 4, only the true 
length of the line B W of the plan Fig. 838 was taken as 
explained from the oblique elevation E' to 4' and placed 
in Fig. S39, 1 7 to E. 

This is the system followed throughout: that for the 
dotted lines being the same and four diagrams, Figs. 839 
to 842 were used to avoid a bewildering mass of fines. 
In all four diagrams, the distances, like 1 7 to 2 6, 17 
to 3 5 and 1 7 to 4 are taken from the circular profile as 
10 to 10°, 9 to 9° and 8 to 8°. The space in all four 




Fig. S44- — The Full Pattern of the Transformation Piece. 

diagrams like E to E*, D to D* in Fig. 839 are taken from 
the plan of Fig. 838 measuring always from the line 
A B, as shown. 

The seam will be set at A of the plan Fig. 838 and 
the oblique elevation of Fig. 838 gives the base lines of 
the solid lines from A around to 1, like A' to I', of the 
oblique elevation, B' to I', C to 2', etc., and the con- 
necting dotted lines as I' to C, 2' to B' etc. In Fig. 843 
is given the base lines for solid and dotted lines from I 
around A; like I to 7 (Fig. 843) J to 8, K to 9, etc., and 
connecting dotted line I to 8, J to 9, K to 11, etc. 

The pattern would be developed by placing the line 
anywhere equal in length to A 1 1 7 of Fig. 839, as 1 A of 
Fig. 844. From 1 in Fig. 844 describe a small arc the 
radius of which equals 1 7 to B* of Fig. 839. From A 
of Fig. 844 describe an arc equal to the length to A°° B°° 
of the pattern Fig. 838. This arc is to intersect the one 



486 



The New Metal Worker Pattern Book. 



from A in Fig. 839 and locates point B. Again from 1 
of Fig. 839, describe an arc of a radius equal to 1 7 C x of 
Fig. 840. Intersect this with one from B of Fig. 844 
the radius of which equals space B°° C°° of the pattern 
of Fig. 838 giving this point C in Fig. 844. From C in 
Fig. 844 describe an arc the radius of which equals 2 6 
C x of Fig. 838. From 1 in Fig. 844 describe an arc to 
intersect the one just made from C, the radius of this 
arc is to equal the division space of the circular profile 
1 to 2 of the oblique elevation, Fig. 838. Continue in 
this manner up to line 7 1 of Fig. 844 when the true lengths 
of lines are taken from Fig. 841 for the solid lines and Fig. 



842 for the dotted lines. Also be sure to take the radius 
of the small arcs of the lettered points from the pattern 
Fig. 838 as directed and the larger spaces, numbered, 
from the circular profile Fig. 838. The final line l x A x 
of Fig. 844, is the same length as the first line 1 A. 

Drawing a line through points 1 to l x and A to A x 
of Fig. 844 results in the net pattern for the transformation 
piece. When shaping objects like these it is necessary 
to be careful to have the right side out of the patterns, 
so that the elbow will offset in the right direction. As 
the patterns are now stretched out the side up will be 
the inside of the objects. 



PROBLEM 246. 



Pattern for the Revolving Blade of a Spiral Conveyor. 



Boilers of heating plants are frequently supplied 
with fuel by means of automatic stokers fed with pul- 
verized coal by spiral conveyors. The conveyor here 
described consists of a pipe in which revolve three spiral 
blades which cany the material forward. These blades 
are identical in construction, and a method is here given 
for developing the pattern for the blade. 

The conveyor pipe is 6 inches in diameter. The 
blades are three in number, and meet in the center, as 
shown in the cut. Each blade makes one-half of a com- 
plete turn in conveying the material two feet (which is 
the usual length of the pipe) as shown in Fig. 845. 

It is impossible to develop a true pattern for a spiral, 
as a certain amount of stretching of the metal is necessary 
to form a spiral from the flat surface. The surface of a 
spiral is a warped surface, and the only pattern which 
would be correct would be a sec ion of a solid. This 
will at once be apparent if an attempt is made to lay 
the spiral out flat after it has been once formed into 
shape. 

Assuming that it is desirable to make blades for a 
small number of conveyors in the beginning for use as 
models, and also assuming :hat they are to be made by 
hand and that the amount of metal used is not as impor- 
tant as the amount of stretching necessary to put it into 
the proper shape, the process for getting the approximate 
rjattern is as follows : 

Draw any line 7 A and with as center draw a circle 
of the diameter of the spiral, in this instance 6 inches. 
This spiral is required to make one-half revolution in the 



length of 24 inches. Draw the line 1' T' parallel to 7 A. 
From 7 through 1' draw a line to 1 and from A through T' 
draw a line to T, making them the length of the half 
revolution, in this case 24 inches. Draw the line 1 T 
in the elevation. The length of the lines 1 T and 1' T' 
in the elevation and the line 7 A in the plan should be 
equal. Divide the arc which represents the one-half 
revolution in the plan into any number of equal parts. 
Divide the length of one-half revolution in the elevation 
into the same number of equal parts. 

Using one of these spaces, B C in the plan, as the 
base B' C and the space 1' 2' from the elevation as the 
hight A' B' construct the triangle A' B' C, as shown in 
Fig. 845. Draw a line half way between A' and B', 
cutting the triangle parallel to the base C D' as shown at 
M and Z. 

Upon any line as D X, Fig. 846, locate the points D' 
and X', using the length of the radius 7 from the plan 
as the distance to space them apart. Through D' and X' 
draw line at right angles from the line D X. Set off the 
distance A" D' equal to A' M and the distance X' X" 
equal to the distance A' Z. Now draw a line through 
A" X" and continue it until it intersects the line D X at 
O' . With 0' as center draw the arc through the point A" 
and continue it on either side, as shown. This arc does 
not pass through D', but slightly above it. With dis- 
tance D' C" equal to the space between A' C, step off as 
many distances on the arc that passes through A" as 
there are spaces in the required rise. Half of the spaces 
are stepped off on each side of the line D X in Fig. 846 for 



Pattern Problems. 



487 



convenience but in actual work they may be all stepped off 
on one side if desirable. 

The line 1° X"' is located at the extreme end of the 
pattern, the arc cutting X" is drawn and continued on 
either side as shown, cutting the extreme end of the 
pattern at X'". As the true length of the pattern on the 
inside edges and the outside edges is only correct after 
the stretching and forming operation, the length of the 
pattern must be determined in the center. 



U' V construct the triangle R' U' V. With the space 
R" U" in Fig. 846 equal to the space R' U' in Fig. 845, 
step off the required number of spaces as shown in Fig. 846 
which will represent the true length of the center of the 
pattern. Joining this line to the point 0' and running it 
up to the edge completes the pattern. 

By repeating the operation on all lines the amount 
of stretching required will be shown. 

The completed drawing of the spiral is not necessary 




Fig. 845. — Preliminary Fig. 848. — Connect- 
Drawiug. ing the Blades. 



Fig. 847. — Straight Pattern. 



Continuing, divide the line 7 in the plan into any 
number of equal parts as shown at U V W. Proceeding 
again in this fashion, divide the line 1° X'" in Fig. 846 
into the same number of equal parts. 

Draw concentric circles in the plan cutting these 
points and draw concentric arcs cutting the correspond- 
ing points in Fig. 846. 

With the space R V in the plan as the base R' V, 
Fig. 845, and the space 1' 2' in the elevation as the hight 



to get the pattern and is given here only to show the ap- 
pearance of the completed article. 

If the triangle T" B" S is constructed with the 
distance B" S made equal to one-half the diameter of 
the spiral and the distance T" B" made equal to the hight 
of the half revolution T T' in the elevation, the dis- 
tance between T" and S will be the true length of the 
outer edge of the finished spiral. And also if the triangle 
so determined is cut out and formed up to the diameter of 



488 



The New Metal Worker Pattern Book. 



the spiral in half circle, keeping the edge T" B" parallel 
to the forming rolls it will serve as a guide for putting the 
spiral partition into shape. 

If these conveyers were to be made in large quantities, 
the partitions could be made of a piece of metal straight 
on all sides, size 6 inches wide and 24 inches long with 
laps added as shown in Fig. 847, in which case the metal 
used would need to be of a thicker gauge and a suitable 
method arranged for drawing and stretching it into the 
proper shape. Fig. 848 shows a suggestion for fastening 
the partitions together with rivets after they have been 
formed into shape. 

If one-half the circumference of the concentric circles 
in the plan which represent lines of the surface of the 
spiral are set off as shown at B" S'", and B" S" and 
B" S' in Fig. 845 and these points joined to T", the 
distances T" S'", T" S" and T" S' will be the true 
length of the corresponding lines on the surface of the 
finished spiral. Divide the body of the pattern in Fig. 
848 into the same number of equal parts as the fines 7 
in the plan and set off the true length of the lines as just 
determined in Fig. 845. Tins shows the amount of 
stretching that will be required at each line. Project the 
line A' to P, Fig. 845, and if all of the work has been 
accurate thus far, the distance P S will be the same as 
the distance A' C Likewise the distance P' S', P" S" 
and J'" S'" will be the same as on the other triangles 



which could be constructed similar to the triangles 
A' B' C and R' U' V in Fig. 845. 

The lines of the pattern in Fig. 846 and the corre- 
sponding fines of the pattern in Fig. 847 should measure 
the same length to the points that are located to show the 
amount of stretching required. 

Now, by comparing the difference between the true 
length of these fines in Fig. 847, it will be easy to see that 
they are unequal or in other words the amount of stretch- 
ing required for lines located equal distances apart from 
the center of the spiral does not vary in equal proportion. 
And tins proves the statement that the surface of a 
spiral is a warped surface. 

If it is desirable to stamp these partitions out in 
parts and no better means is available, it may be done by 
making a solid pattern for the sweep of the desired section, 
using plaster of Paris, and painting with one or two coats 
of shellac. 

This may be sent to an iron foundry and the pattern 
for the sweep cast and then a lead form moulded at the 
shop from the iron casting or, instead of an iron casting, 
scrap zinc is suitable for use with lead, but of course would 
not do for long service. 

With the aid of a drop weight press or even a heavy 
sledge and by putting the metal between the upper and 
lower forms just described, the parts may be stamped 
out. 



PROBLEM 247. 



A Collar Intersecting a Transformer. 



The pipe and branch shown in Fig. 848 A furnishes 
a very interesting and intricate problem. It is necessary 
to get out patterns for a 7-inch collar intersecting a 
transformer at an angle of 60 degrees, with the dimensions 
as given. The problem is to develop a pattern for both 
the collar and the opening to be cut in the transforming 
section. It will be noted that the 8-inch round end of the 
transformer in the perspective view is divided into twelve 
spaces, and that the axis of the collar would intersect at 
about the point 2, thus apparently setting on the corner 
of the transformer. In the elevation this collar sets at 
an angle of 60 degrees 3 inches from the corner of the 
transformer. In the plan the transformer sets in a 
central position. 



In this case the conditions are seemingly impos- 
sible of fulfillment. In the diagram, Fig. 849,' is a 
reproduction to scale of the problem. Analyzing this 
problem on a practical basis, it must be said that it 
is a very queer proposition, and it is wondered if a 
more simple solution would be possible, if all the con- 
ditions were known, viz.: What is the object? Where 
does it go? And isn't there a run of 8-inch pipe joined 
to the transformer, and if so, why couldn't the 7-inch 
collar be tapped into the 8-inch pipe with an ordinary 
tee joint or Y, or, perhaps, there is a rectangular pipe 
joined to the square part of the transformer and pos- 
sibly the 7-inch collar could be tapped into it. And, 
again, isn't it unusual to have the miter line B as given, 



Pattern Problems. 



489 



instead of the customary miter line B', which bisects 
the angle C. 

Note that the profile of the 8-inch end is divided 
into the number of parts directed. Also that axis A' of 
the 7-inch collar is projected down to intersect No. 2 
element at A at the transformer as directed, and would 
be point A" when projected down to the plan. A par- 
tial plan view of the 7-inch collar is shown, and it 
will be seen that a large part of the collar falls clear of 
the transformer, or positively will not intersect the trans- 
former. 

Then wnat does this mean? It means that the 
requirements laid clown cannot be fulfilled, and therefore 
some modification of the conditions is necessary. Shall 
the collar be made a sort of offset elbow so that all of it 



v.^ 



r -/3r-i 




Perspective 



Plan 
Fig. 84S A. — The Problem as Presented. 



will intersect the transformer? That would be one 
method, and would seem to be the best plan to follow in 
this particular case. 

The elevation, shown in Fig. 849, is drawn to the 
specifications furnished. Note the positions of the 
profiles of the 7-inch and 8-inch collars, the 8-inch collar 
being spaced as shown. The plan is made by drawing a 
square to the dimension of that part of the object. Center 
line P X and its points are placed in the plan as P' X' and 
horizontal projector lines from this, intersecting vertical 
projector lines from line Z Z', gives a plan view of round 
end of transformer. Complete the plan as shown. 

To avoid mixing things up too much, that portion 
of Fig. 849 which is really needed is reproduced as in 
Fig. 850. Divide the 7-inch profile into twelve spaces, 
and drop lines to line C B in the elevation. Project 



these points on C B down to plan. Draw the 7-inch 
profile in the plan and project lines across to intersect 
those from line C B. Observe that it is merely coinci- 
dence that lines from 9 and 3 pass through points, as at 
a, of the round end of transformer. It is also coincidence 
that lines 9 and 3 should touch fines C B' and C B". 




Plan 



Fig. 849. — Reproduction of the Problem to Scale for Analysis. 

It will be noticed now that a major part of the 7-inch 
collar intersects that flat part C B' C B" of the surface 
of the transformer, and that but one set of elements of 
the collar, 2 and 10, touches the rounded surface of the 
transformer. Of course, if the collar were made to 
offset both ways, that is, if it were offsetted so that all 
of the collar fell on the flat surface, a much simpler miter 
fine would be obtained, but it would complicate the 



490 



The New Metal Worker Pattern Book. 



developing of the collar patterns to the extent of making 
it not worth the trouble. 

So then it is, therefore, necessary to determine just 



projected to the plan to like elements of the transformer, 
giving C" D" E" F". A line traced through these 
points gives the lines of intersection in plan of where 




0-F 



1 --'2 "3 -4 
Fig. 851. — Length of Elements. 



Fig. 853. — Half Pattern of Square Part of Transformer. 



where these elements touch the mentioned rounded sur- 
face. Element or projector 2 and 10 in elevation is 
drawn, or rather continued, to pass through elements 
C D E F. Where they intersect, as C D' E' F', they are 



elements 2 and 10 intersect those, in elevation, of trans- 
former elements. Where point 10 in plan intersects 
this line (or the same thing point 2), as 10', is where 
these elements, 2 and 10, touch the rounded part of the 



Pattern Problems. 



491 



transformer. This point 10' is projected up to inter- 
sect line 2 10 in elevation, giving point 10". To facilitate 
matters, another point is introduced where the line of 
intersection of the collar leaves the flat surface of the trans- 
former as A' and A" and A and A'" on profile. This is 
projected up to the elevation as A"". A line traced 
through points in elevation as shown gives sought-for 
line of intersection, or as geometricians term it, line of 
penetration. 

An oblique view is now drawn of the collar to delin- 
eate amount of necessary offset. Draw line 6 6' parallel 
to the collar in elevation. Also draw line K parallel to 
6 6', and as far from 6 6' as A" K in plan of Fig. 849. 
Draw 7-inch collar profile, as shown, being careful to have 
numbered points situated correctly, especially point A, 
being guided by profile in plan. 

Draw the offset piece R to what would seem about 
the right size. Project lines up from the profile in the 
oblique view to terminate at miter line S. Also intersect 
these hnes with projectors from the elevation. Be careful 
to select correct points in the elevation, especially A"", 
then 10" and then 3 and 9. Observe the contour of the 
cut in the collar in oblique view, at the intersecting 
end of the collar. 

Draw a line as 1 A 2 3, etc., spaces being taken 
from the 7-inch profile. Draw lines through these points 
as shown, which are intersected by hnes from the oblique 
view — to 0; 1 to 1; A to A, etc., both from miter line S 
and the bottom cut of the collar, which gives the pattern 
of piece T. Set the dividers to span h h of oblique view 
of piece R, then step this off on the stretchout lines of 
patterns, as h' h' at 9, which is then the pattern of piece R. 
For the pattern of piece W take the distance h m in the 
oblique view and place it on stretchout fine 9 in the pat- 
tern, as h m. Parallel to the stretchout line, draw a line 
through to, which will then give the pattern for piece W. 
When joining these three pieces the X marks should be 
opposite each other. 

For the transformer pattern it is first necessary to 
find out actually how long the various elements are; 
so in Fig. 851 draw a line and place thereon the lengths 
of the elements as given in the elevation of Fig. 849, 
for instance, 6' F in elevation, Fig. 849 is F 6 in Fig. 851, 
and say, 3' C in Fig. 849 elevation is C 3 in Fig. 851. 
Erect perpendicular hnes from these points. Make C C 



of Fig. 851 equal to space C C" in plan of Fig. 849. 
Make fine 2 equal to 2' 2 in 8-inch profile in elevation of 
Fig. 849, and so on, which gives the true lengths of the 
elements. In Fig. 852 draw a fine, F F' equal to C C" 
of plan Fig. 849. At right angles to F F', in Fig. 852, 
draw line F' 7 equal to F Z' or 7 of elevation, Fig. 849. 
Join 7 to F; then with F as center, swing short arcs of 
radii respectively of Fig. 851, C 6', C 5' and C 4'". 
Set the dividers to place the 12 spaces of the 8-inch pro- 
file in elevation, Fig. 849, and starting at 7, in Fig. 852, 
step on the arcs to 4. 

From 4 as center, in Fig. 852, draw an arc of a radius 
equal to 4" C cf Fig. 851. Intersect this arc with one 
from F, in Fig. 852 of a radius of C F in elevation of Fig. 849 
which is miter fine B of Fig. 849. This establishes point 
C in Fig. S52, and with that as center, describe short arcs, 
taking their radius from Fig. 851, as C 1, C 2, etc. Step 
these off, as from 7 to 6, giving then points 3 2 and 1. 
From 1, as center, describe arc of a radius equal to 2 C 
of elevation, Fig. 849. Intersect this with an arc of a 
radius equal to F' F' of Fig. 851, giving point C, which 
completes the one-half pattern of the transformer. 

For the hole to be cut out for the collar, take the 
points in fine C B, in Fig. 850, and place them on fine 
1 C of Fig. 852, as 1 to 0, 1 to 1' and 11, 1 to A, omitting 
10 and 2, and then 1 to 4' and 8", etc. Draw hnes at 
right angles to 1 C and then place on 1' and 11' the dis- 
tance from center line C in plan, Fig. 850 to 1*. Then 
on A, in Fig. 852, fine C to A' in plan, Fig. 849, and so 
on. Now, the best way to find the location of point 10 
or 2 — which is on the rounded part of the transformer — is 
to set the compass to the space A x 2 X of the pattern of 
offset in Fig. 850, and then using A 1 , in Fig. 852, as 
center, swing a small arc. Set compass now to space 
T 3 1 in the pattern of offset, Fig. 850, and with 3*, in 
Fig. 852, as center, intersect the arc from A x , locating 
thereby point 2 X . Trace a hne through these points, as 
shown by the cross-hatching, which is the outline of the 
hole for the collar, or line of penetration. 

Fig. 853 is the one-half pattern of the square part of 
the transformer, and is simply the placing of space C C", 
in the plan of Fig. 849, on a line, as shown. From C" to 
C is double the distance of C C" and C" to P is C P, in 
Fig. 849 elevation, while C D, in Fig. 853 is F D in eleva- 
tion Fig. 849. 



492 



The New Metal Worker Pattern Book. 



PROBLEM 248. 



Patterns for Cone Hopper Intersected by a Spout. 



As will be noticed, especially by Fig. 854, the hopper 
is round, of conical shape, and is intersected by a rect- 
angular tapering spout, the taper of which is in the side 
elevation only, the front of the spout showing parallel 
lines, as indicated in the front elevation. This model 
shows the one side of the spout a b in the front elevation 
to be in line with the center line of the conical spout, while 
the tapering side of the spout is to have the position 
shown in the side elevation. The problem presented 
by our correspondent gives an interesting study in pro- 
jections, intersections and developments, and is worked 
out as shown in Fig. 855. 

First, draw the center line A B, on either side of 
which draw the cone hopper as shown by 4 4 B in eleva- 



r Round Hopper ... 





Rectangular h H 

—Tapering Spout-" ^ 



Side 
Elevation 



Front 
Elevation 



Fig. 854- — Reproduction of Paper Model. 



tion. Locate as desired, the center line of the spout as 
indicated by a b. Establish the hight a b, also the semi- 
width at the top a 7 and a f, as well as the semi-widths 
at the bottom, as b d and 6 e. Now draw the taper 7 d 
and / e. 7 / e d then represents the side elevation of the 
spout which is to intersect the cone hopper. Above the 
line 4 4 in elevation draw the plan of the hopper, shown 
by 4 C 4 D. 

As the spout is to set to one side of the center of the 
cone, as shown by a b in Fig. 854, and as the sides are to 
be parallel as shown in this view, then establish the 
desired width as indicatedby X in plan in Fig. 855, and 
draw a line parallel to 4, 4, in plan until it intersects 
the center line A D in plan at c' and the top outline of 
the cone at 4'. As the one side of the spout sets on the 
center line 4 4 in plan, then will 7 8 e d in elevation be 



the pattern for that side. To obtain the pattern for the 
side on the line 4' c' in plan, it will be necessary to obtain 
a vertical plane of the cone on the line 4' c' in plan as 
follows. Using A as center with radius equal to A c' 
draw the semi-circle cutting the center line 4 4 and 1 1. 
From point 1 on the right side in plan, at right angle to 
4 4, drop a line intersecting the right side of the cone 
in elevation at 1, from which point at right angles to the 
center line A B draw the line 1 1, which represents a hori- 
zontal plane shown in the semi-plan by 1 c' 1. Divide 
the distance between 1 and 4 on the left side in plan in 
any convenient number of spaces (in this case three), as 
shown by 2 and 3. 

Then, using A as center with radii equal to A 2 and 
A 3 draw the semi-planes 2 2 and 3 3, intersecting the 
side of the spout 4' c' in plan at 2' and 3'. From points 
2 and 3 on the right side of the center line 4 4 in plan 
drop perpendiculars, cutting the side of the cone at 2 and 
3, from which points draw similar horizontal planes in 
elevation, indicated by 2 2 and 3 3. From the inter- 
sections 2' and 3' in plan drop perpendicular lines in the 
elevation cutting similar planes 2 and 3 at 2' and 3'. As 
the distance between the points c' and 2' in plan is too 
great, bhect this distance and establish another point as 
11' and using A as center, and radius equal to A, 11' draw 
the semi-plane 11 11. 

Project point 11 on the right-hand side to the eleva- 
tion, cutting the side of the cone at 11, and draw the 
horizontal plane in elevation as 11 11, which intersect at 
11' by a perpendicular line dropped from the intersection 
in the side of the spout 11' in plan. From 4' in plan 
project a perpendicular to the top line in elevation, also 
shown by 4'. Through the points 4' 3' 2' 11' c' in eleva- 
tion trace the curved line, as shown, which represents 
a vertical plane in elevation on the line 4' c' in plan. 

Now extend the sides of the spout in elevation as d 7 
, and e 8, cutting the plane in elevation at 6' and 5', respec- 
tively. Then will 6' 5' e d represent the pattern for the 
side of the spout on the line 4' c' in plan. Project the 
points 5' and 6' in elevation up to the plan, thus locating 
similar points shown by 5' and 6'. From points 5' and 6' 
in elevation draw horizontal lines cutting the sides of the 
cone also at 5 and 6. From these two points 5 and 6 on 
the right side of the cone B 4 erect perpendicular lines 
to the plan cutting the center line 4 4 in plan at 5 and 6. 
Again using A as center with radii equal to A 5 and A 6, 



Pattern Problems. 



493 



draw the semi-planes 5 5 and 6 6, which intersect by lines 
drawn vertically from 5' and 6' in elevation at 5' and 6' 
in plan, as shown. 

It will be noticed that these two lines just projected 
from 5' and 6' in elevation come directly on the inter- 
sections between the planes 5 5 and 6 6 in plan 
and the side of the spout 4' c', thus proving 
the operations. So that an accurate pattern for 
either the right or left side of the spout can be 
obtained, establish at pleasure any point as 9' 
between 5' and 8 in elevation and between 6' 
and 7 in elevation as shown by 10'. 

Through these two points 9' and 10' draw 
the horizontal planes in elevation as indicated 
by 9 9 and 10 10. From 9 and 10 on the left 
side of the cone erect perpendicular lines to the 
plan cutting the center line 4 4 at 9 and 10, re- 
spectively, and using A as center with radii 
equal to A 9 and A 10 draw the short arcs 
shown by 9 9 and 10 10 in plan, which inter- 
sect by vertical lines erected from points 9' and 
10' in elevation, thus obtaining the intersections 
9' and 10' in plan. Project the points 7 and 8 in 
the elevation of the spout, to the plan, thus 
locating these two points in plan, all as shown 
by the dotted lines. 7 6' 5' 8 in plan then 
shows the plan view on 7 6' 5' 8 in elevation, 
while h i n 3' shows the plan view of the bottom 
of the spout on d e in elevation. For the pat- 
tern for the left side of the spout proceed as fol- 
lows: At right angles to 6' d in elevation from 
points 6', 10', 7 and d draw lines indefinitely as 
shown. Parallel to 6' d draw the line 6 X G. 
Take the width of X in plan and set it off from 
6 1 to 6" and G to H and draw a line from 6° to H 
as shown. Now measuring in each instance 
from the line 4 4 in plan, take the vertical dis- 
tances to points 10' and 6' and place them in 
the pattern on similar numbered lines, measuring 
in each instance from the line 6 3 G, thus obtain- 
ing the points of intersections 10° and 6° as 
shown in the diagram. 

Trace the curved line 6" 10" 7*. Then will 
6° T G H be the desired pattern. For the pat- 
tern for the right side, draw lines indefinitely from points 
5' 9', 8 and e at right angles to 5' e, parallel to which 
draw the fine 5 X J. Make the distance 5 X 5° and J K 
equal to X in plan, and connect the two points 5° and K. 

Now, again measuring from the line 4 4 in plan 
measure the vertical distances to points 9' and 5' and 



place them on similar numbered lines in the pattern, 
measuring in each instance from the fine 5 X J as shown 
from 9* to 9' and 5 1 to 5". Trace the curved fine 5", 9", 8 Z . 
Then 5' K J 8* is the desired pattern. 

If the pattern is desired in one piece, then the various 




Fig. 855. — Plan and Elevation, with Pattern Shapes Developed. 

patterns can be joined together as follows: Take a tracing 
of the pattern for the right side of spout, indicated by 
8 T J K 5° and on the fine 5° K join the pattern of the taper 
side, shown in elevation by e 5' 6' d and to the side 6' d 
join the pattern for the left side, as indicated, by 6° H G T 
and on the line 7 1 G join the other tapered side indicated 



494 



The New Metal Worker Pattern Book. 



by 7 d e 8, all as shown complete in one pattern in Fig. 
856. To obtain the shape of the opening to be cut in the 
side of the conical hopper, proceed as shown in Fig. 855. 
Using B as center with radii equal to B 8, 9, 5, 11, 2, 3, 7, 




Fig. 856. — Pattern for One-piece Spout. 

10, 6, and 4 draw the arcs as shown. At pleasure from the 
apex B draw any line as B E. 

Now, measuring from the line 4 4 in plan, measure 
off along the curve from 6 to 6' and place it on the arc 
6 6 in the pattern measuring from the line B E along the 
curve from 6° to 6'. 



In a similar manner measuring from the line 4 4 in 
plan take the distance along the curve from 10 to 10' and 
place this distance along the curve 10 10 in the pattern 
from 10° to 10'. 

As the one tapering side in elevation lies directly on 
the side of the cone and intersects the various planes in 
elevation at 7, 3, 2, 11, 9, and 8, then will the arcs struck 
from the apex B and intersecting the line B E at 7°, 3°, 2°, 
11°, 9°, and 8° show the true points of intersections and a 
line drawn from 8° to 7° in the pattern shape of opening 
will be the outline on one side. Now in a similar way, 
measuring in each instance from the line 4 4 in plan take 
the various distances along the curves or arcs, as from 
3 to 3'; 2 to 2'; 11 to 11'; 5 to 5', and 9 to 9', and place 
these distances in the pattern measuring in each instance 
from the line B E, along the curves having similar num- 
bers as shown respectively from 3° to 3'; 2° to 2'; 11° 
to 11'; 5° to 5' and 9° to 9'. Trace a line through points 
thus obtained as shown from 8° to 5' to 6' to 7°. Then 
will the shaded portion represent the shape of opening to 
be cut in the conical hopper. The full pattern for the 
conical hopper is not given as that is simply the pattern 
for a right cone in which the shaded opening must be 
cut. 



PROBLEM 249. 
Pattern for Gooseneck with Compound Curve. 



This discussion is on methods of how to obtain 
the patterns for a gooseneck having a compound double 
curve as shown in Fig. 857, in which the plan, front and 
side elevations are shown. In the plan, A represents 
the gutter from which a gooseneck is to connect at B, 
and pass over to the inside corner of the wall at an angle 
of 45 degrees, as indicated by C. The front elevation 
of this gooseneck is shown by D, the projection from the 
wall being equal to c a and hight equal to a b. In the 
elevation E shows the side of the gooseneck, the projec- 
tions a' c' and a' v' being similar to those in the front view. 

The usual method of developing these patterns is by 
parallel lines, but in Fig. 858 is shown how the patterns 
can be laid out by radial lines, and the area of the pipe will 
be maintained throughout the entire curve. 

In Fig. 858 let A B C D represent the section of the 
top of the leader at the gutter and A 1 B 1 C 1 D 1 similar 
section on the inside corner of the wall. Let the pro- 
jection between the gutter and wall be as much on either 



side as is indicated by the arrows X and Y. Connect 
opposite corners of the leader sections as shown from A to 



Wall line 



Wall line 

- ' - ' --~'i&" ' s J- \- - -{■ ■ > ^~i'; 

m 




Fig. 857. — Elevations in Gooseneck in Problem. 



Pattern Problems. 



495 



A 1 , B to B 1 , C to C 1 and D to D 1 . At right angles to 
these lines an oblique view must now be constructed as 
follows: 

Parallel to the line A A 1 in plan draw the line 
E F and G H, making the distance between these two 
lines equal to the hight of the gooseneck. From the 
corners in the section A B C D erect lines at right angles 
to A A 1 , cutting the top line G H at A, B, C and D, as 
shown. In a similar manner erect lines from the sec- 
tion A 1 B 1 C 1 D 1 in plan, cutting the base or bottom line 
E F at A 1 , B 1 , C 1 and D 1 . 




proceed as follows: At right angles to G H from the point 
J, draw the line 3 m. In a similar manner at right angles 
to E F from the point R, draw the line K n. Extend the 
lines D A, A B, C B and D C in the section in plan until 
they intersect the line projected from the center J in the 
oblique view at d, e, f and h, respectively. In a similar 
manner extend the lines A 1 D 1 , B 1 A 1 , B 1 C 1 and C 1 D 1 
until they intersect the line projected from the center K 
in the oblique view at d', e', f and h', respectively. These 
various points d, e, /, and h and d', e', /', and h' represent 
the centers with which to develop the patterns for the 
sides of the elbow. 

The next step is to divide the various curves in the 
oblique view into equal spaces, from which the girth is 
obtained in developing the patterns. Thus, divide the 
outer curves from D to D 1 in the oblique view, as shown 
from 1 to 10 and the inner curve B 1 to B, as shown from 
11 to 20. 

To obtain the pattern for the side of the gooseneck 
from D to A in plan proceed as follows: Draw any hori- 



C B 




50 

\9 s / 

va 



. Pattern for 
ITK - >■ SideB-C 




Fig. 858. — Method of Developing Pattern for Gooseneck with Compound Curve by Radial Lines. 



At pleasure establish the centers J at the top and K 
at the bottom with which to describe the curves shown. 
The radius J B at the top and K D 1 at the bottom may be 
made similar or dissimilar as desired. 

Using J as center and with radius equal to J B draw 
the arc B t. With radius equal to KB 1 at the bottom 
draw the arc B 1 u. Tangent to these two arcs, draw the 
line 16 17 and from J draw a line through 17, extending it 
indefinitely. In a similar manner from K draw a line 
through 16 indefinitely as shown. With radii equal to 
J C, A and D draw arcs cutting the line J 6 as shown. 
Also with radii equal to K C 1 , A 1 and D 1 draw arcs cutting 
the line K 16 as shown. Connect arcs by straight lines 
as shown. Then B D D 1 B 1 represents the oblique view 
of the gooseneck. 

To obtain the radii with which the patterns are struck 



zontal line in diagram as shown by D d. With radii 
equal to d A and D in plan and d in as center draw 
the arcs D 6 and A 6'. On the arc D 6 set off the girth of 
1 to 6 in the oblique view as shown from 1 to 6 in and 
from 6 draw a radial line to d, cutting the inner arc at 6'. 
At right angles to the radial line 6 d draw lines from 6 and 
6' equal in length to 6 7 in the oblique view as shown in 
by 6 7 and 6' 7'. Extend the line T 7 indefinitely as 
shown. With radius equal to d' A 1 in plan and with 7' in O 
(which represents the fine of the corner joint A or A 1 ), 
draw the arc d' intersecting the lines drawn from 7'. 

Using the same radius, with d' as center draw the 
arc 7' A 1 . Again using the center d', with radius equal 
to d' 7 draw the arc 7 10. Take the girth from 7 to 10 
in the oblique view and place it as shown from 7 to 10 in 
diagram O, and draw a line from d' through 10 intersecting 



49G 



The New Metal Worker Pattern Book. 



the outer curve at A 1 . Then ADD'A 1 represents the 
pattern for the side of the gooseneck shown by A D or 
A 1 D in plan. 

To obtain the pattern for the side A B in plan, draw- 
any line, as A e in diagram P. With radii equal to c B 
and A in plan and e in P as center, draw the arcs B 17 
and A 17' respectively. On the arc B 17 set off the girth 
of 20 to 17 in the oblique view, as shown by similar num- 
bers in P. Draw a hue from e through 17 cutting the 
outer arc at 17'. From points 17 and 17' draw the lines 
17 16 and 17' 16' at right angles to e 17' equal to 17 16 in 
the oblique view. Extend 16 16' indefinitely as shown in 
P. With radius equal to e' B 1 in plan, and with 16 in P 
(which represents the line of the corner joint B or B 1 in 
plan), describe the arc e' intersecting the line previously 
drawn from 16. With the same radius and e' as center 
draw the arc 16 B'. Again using e' as center and e' 16' 
as radius describe the arc 16' A 1 . Take the girth from 16 
to 11 in the oblique view and place it as shown from 16 to 11 
in P, and draw a fine from 11 to e' cutting the inner arc 
at A 1 , and A B B 1 A 1 will show the pattern for the side 
A B or A 1 B 1 in plan. 

The pattern for the side D C in plan is obtained 
by drawing any line, as D h in diagram R. With radius 
equal to h C and D in plan and h in R as center draw the 
arcs C 6' and D 6. On the arc D 6 set off the girth of 1 
to 6 in the oblique view, as shown by similar numbers in 
R, and draw a line from 6 to h intersecting the inner 
arc at 6'. At right angles to 6 h from points 6 and 6' 
draw the lines 6 7 and 6' 7' equal in length to 6 7 in the 
oblique view. Extend 7' 7 indefinitely as shown, and 



using h' C 1 in plan as radius and with 7' in R (which 
represents the fine of the corner joint C or C 1 in plan), 
describe the arc h' intersecting the fine drawn from 7'. 
With the same radius and h' as center describe the arc 
7' C 1 . Again using h' as center and h' 7 as radius, 
describe the arc 7 D 1 . Take the girth from 7 to 10 in the 
oblique view and place it as shown from 7 to 10 in R and 
draw a fine from h' through D 1 until it cuts the outer arc 
at C 1 . Then D C C 1 D 1 is the pattern for the side 
C D or C 1 D l in plan. 

For the last pattern draw any line, as / C in S. 
With radius equal to / B and C in plan and / in S as center 
describe the arcs B 17 and C 17'. On the arc B 17 set off 
the girth of 20 17 in the oblique view, as shown by similar 
numbers in 3. From / draw a line through 17 until it 
meets the outer arc at 17'. From 17 and 17' draw fines 
at right angles to 17' / equal in length to 17 16 in the 
oblique view. Draw a line in S through 16 and 16' 
indefinitely as shown, and using/' B 1 in plan as radius and 
16 in S (which represents the corner joint fine B or B 1 
in plan) as center, describe the arc /' intersecting the 
fine drawn from 16. Use the same radius and/' as center 
and draw the arc 16 11. Again using/' as center and/ 16' 
as radius, describe the arc 16' C 1 . Take the girth from 
16 to 11 in the oblique view and place it on the outer 
curve in S, as shown by similar numbers, and draw a 
line from 11 to /' cutting the inner arc at C 1 . Then 
BCC'B 1 shows the pattern for the side B C or B 1 C 1 
in plan. 

Laps must be allowed for soldering, seaming or rivet- 
ing, according to the method of constructing. 



PROBLEM 250. 
Short Rule for Spiral Strip in Separator. 



The following text and diagrams describe a short 
method of developing the spiral strip in a dust separator 
without the use of triangulation; it, however, must 
be remembered that only an approximate pattern results 
inasmuch as no method can develop a strictly accurate 
pattern. Referring to Fig. 859 proceed as follows: 

Divide the outline of the plan in an equal number 
of spaces, in this case 8, as shown by the small letters 
a, b, c, d, e, /, g and h, from which points draw lines to 
the apex r. Assuming that the spiral strip is to make 
two revolutions, divide the vertical bight of the sepa- 
rator in two tunes the number of spaces in the plan, or 



2X8 or 16 spaces, as shown from a' to h' on the vertical 
fine a' t. From the divisions a to h in plan, erect fines, 
partly shown, intersecting the top line of the separator 
as shown by similar letters, from which points radial 
lines are drawn through the elevation to the apex X. 

Now from similar lettered divisions on a' t, draw hori- 
zontal lines intersecting similar radial fines in elevation 
at 1, 2, 3, 4, 5, etc., to 17, as shown. 

If desired trace a spiral fine through these divisions 
as shown. From these numbered intersections 1 to 17 in 
elevation, drop perpendicular fines in the plan, cutting 
similar lettered radial lines as shown by similar points of 



Pattern Problems. 



497 



intersection 1 to 17. If desired trace the plan view of the 
spiral through points just obtained, although this is not 
necessary in the laying out of the pattern. 

The pattern strips are now in order and are developed 
by means of parallel lines. The method to be given is 
based on the principle that a cone cut by a plane obliquely 
through its opposite sides as A B in elevation produces 
an elliptical figure. 

Therefore for the pattern for the upper revolution of 
the spiral proceed as follows: Take the girth in elevation 
of the spiral line, 1, 2, 3, 4, 5, 6, 7, 8, ana 9, and place it 
on the center line in plan extended as e i as shown by 
1, 2, 3, 4, 5, and then go backwards to 6, 7, 8, and 9. 
Through these points draw vertical lines which intersect 
by' fines drawn parallel to e i from similar numbered 
points of intersections in plan. 

Trace the outline 1, C, 5, D, 9, 
which will meet the inside of the cone. 
Now set the dividers to the required 
width as I m and scribe the inner line 
9 m 8 as shown. 

In precisely the same manner obtain 
the pattern for the lower revolution. 
The girth of 9 to 17 in elevation is 
placed on line a n in plan, the usual 
measuring lines drawn and intersected 
by lines drawn from similar numbers in 
plan. Trace the outline of the lower 
pattern as shown, and allow the proper width V m' , 
as shown. In this case the patterns have been made in 
two parts. 

In making a spiral for a full-size separator, one-quarter 



sections would be used, that is, four pieces for every revo- 
lution. As the pattern is but approximate, slight stretch- 
ing of the edges will be necessary, before turning the rivet- 



Elevation 




Pattern for 
Lower Revolution 



Pattern for 
Upper Revolution 



Fig. 859. — Short Method of Developing Pattern for Spiral Strip in Separator. 



ing flange. In riveting in the spiral, best results are 
obtained by starting at the bottom of the cone. The 
patterns are net and riveting flanges must be allowed to 
all patterns. 



PROBLEM 251. 
Pattern for Offset in Ventilation Pipe. 



The following demonstration is an interesting and 
unique procedure for developing a pattern for offset in a 
T ventilation pipe which is 48X72 niches and made of 
16-gauge galvanized iron. 

In Fig. 860 is shown a picture of a finished offset 
which is shown developed in Fig. 861. The first step 
is to draw the profiles of the pipes in plan in their proper 
positions, as shown by 1 2 3 4 and 1' 2' 3' 4', giving the 
proper projection from the line H to the side of the lower 
pipe 2 3 and from the fine G to the side of the pipe 1 2, 
as desired. Through the two profiles draw diagonal fines 



(not shown), intersecting each other at b and a respect- 
ively, and through the two intersections draw the line 
A B. At right angles to this fine a diagonal elevation 
must be constructed as follows: Parallel to A B draw 
the line E F, and at the desired distance as E D draw 
the fine D C parallel to B A. 

From the points 1 2 3 4 in the profile in plan draw 
lines at right angles to A B, cutting the line C D in the 
diagonal elevation as shown by similar numbers. In a 
similar manner, at right angles to A B from the corners 
1' 2' 3' 4' in the profile in plan, draw lines cutting the top 



498 



The New Metal Worker Pattern Book. 

•6 








Fig. 860. 
E Finished Elbow 






V 



\\\^xV 



S-T-U-V Patterns for 
Upper Part of Elbow 

Ft 



\ / Vv 

// / - 

tf /Y / / / 

y / / / 



/ 3 0/ S // 





4 3 4 T 

Fig. 861. — Developing Patterns for Offset in Ventilation Work. 



~0 2 



Pattern Problems. 



499 



line F E in the diagonal elevation as shown. Now, 
establish at pleasure the centers 5 and 6 in the diagonal 
elevation with which the throats of the elbow are struck. 
Using 5 as center on the top line E F, with radii equal to 
5 4', 5 3', 5 1' and 5 2', draw arcs indefinitely as shown 
respectively by c, d, e, and /. In a similar manner with 
any desired eenter as 6, on the line C D, with radii equal 
to 6 2, 6 3, 6 1, and 6 4, draw the arcs shown respectively 
by g, h, i, aad j. Now connect similar numbered arcs 
by fines drawn tangent to each arc as indicated from 4° 
to 4°, 3° to 3", 1° to 1' and 2° to 2". 

From the point 4" draw a radial fine to the center 6 
and from the point 2° draw a radial fine to the center 5. 
Connect the points of tangency by the fines 4° to 3° to 2° 
to 1° to 4°, also from 2" to 3" to 4' to 1° to 2". From these 
points of tangency 1° to 4° and 1" to 4" erect lines in the 
plan at right angles to A B, intersecting similar num- 
bered lines as shown from 1° to 4° and 1" to 4". Connect 
these points as shown, and draw lines from 1° to 3" to 4°, 
also from 1° to 3° to 2". These fines will be used when 
laying out the straight portion between 1° and I s by 
triangulation, while the circular parts of the elbow will be 
developed by radial lines. 

To obtain the radii for striking the various sides of 
the curved elbow, proceed as follows: From the center 
point 5 at right angles to F E draw the line 5 G indefinitely. 
Now extend the sides of the pipe 2' 1', 2' 3', 2' 4', and 
1' 4' until they intersect the line 5 G at G, H, J, and K, 
respectively, which joints represent the centers for strik- 
ing the patterns. In a similar manner at right angles to 
C D in the diagonal elevation from the point 6 erect the 
line 6 indefinitely, as shown. Now extend the sides 
of the pipe 1 2, 3 2, 4 3, and 4 1, until they intersect the 
fine 6 at L, M, N, and 0, respectively. 

The pattern for the side 1' 2' of the upper curve is 
obtained by using G 1', G 2', in plan as radii and G 1 in 
diagram S at center, and describing the arcs V 1° and 
2' 2°, as shown. As the fine 1 V in plan lies in a hori- 
zontal plane or parallel to A B, then will the compound 
curve 1 1' in diagonal elevation show a true section or 
profile from which measurements can be taken and placed 
on the pattern shape being developed in S. Therefore, 
starting from 1 on the lower part of compound curve 1 1' 
in diagonal elevation, divide that part between 1 and the 
radial line 4" 6, as shown by the points numbered 1, 7, 8, 
9, 10, and 11. In a similar manner, starting from where 
the radial fine 5 2° intersects the compound curve at 12, 
divide the distance between 12 and 1' into equal spaces 
as indicated by the numbers 12, 13, 14, 15, 16, and 1'. 

Now take the various divisions between 1' and 12 in 



the diagonal elevation, and starting from 1' in the pattern 
S, step off similar divisions as shown from 1' to 12. From 
the apex G 1 draw a radial fine through 12 and 1', inter- 
secting the outer curve at 2° and 2'. Now take the dis- 
tance from 12 in the diagonal elevation to the point of 
tangency 1°, and set it off from 12 to 1° on the inner arc 
in the pattern S. Draw a fine from 1° to 2°. Then 1° 2°, 
2' V is the pattern desired. Now with radii equal to 
H 3', H 2' in plan and H 1 in diagram T as center, draw the 
arcs 3' m and 2' 2°. Divide the outer arc 2° 2' in dia- 
gram S in equal spaces as shown and place the same 
numbered divisions along the outer arc in diagram T, as 
shown, from 2' to 2°, and from 2° draw a radial fine to the 
apex H 1 intersecting the inner arc at m. Now, measure 
the distance from the point of tangency 3° in the diagonal 
elevation to the point m on the radial fine and place it 
from m to 3° in the pattern T and draw a fine from 3° to 
2°, also a fine from 2' to H 1 , cutting the inner arc at 3'. 
2' 3' 3° 2° is then the pattern for the side 2' 3' in the upper 
curve. Now, using from the plan the radii J 4', J 3' and 
J 1 in diagram U as center, describe the arcs 4' 4° and 3' m. 

Draw a fine from 3' to J 1 , cutting the inner arc at 4'. 
Now, divide the inner arc in pattern T between 3' and 3° in 
equal spaces as shown, and place these divisions along 
the outer arc 3' m in pattern U, as shown by similar 
numbers between 3' and 3°. Take the distance from 3° 
to m in T and place it from 3° to m in U. Draw a fine 
from m to J 1 , cutting the inner arc at 4° and draw a line 
from 4° to 3°. Then 3' 4' 4° 3° will be the pattern for 
the upper part of the elbow whose side is marked 3' 4' in 
plan as shown. 

Using as radii K 4', K 1' in plan and K 1 in diagram 
V as center, describe the arcs 4' 4° and 1' 1°. From 1' 
draw a radial fine to the apex K 1 , intersecting the inner 
arc at 4'. 

Now, take the girth along the inner curve 1' 1° in the 
pattern S and place these divisions along the outher curve 
in pattern V as shown by similar numbers. From 12 
draw a radial fine to the apex K', cutting the inner arc 
at 4°, and draw a line from 4° to 1°. By measuring the 
divisions along 4° 4' in pattern U, they will be found to 
correspond to similar girth along 4° 4' in pattern V. 
The distance 12 1° along the outer arc in V will also cor- 
respond to 12 1° in the diagonal elevation. 4' 1' 1° 4° 
will then be the pattern for the side 4' 1' in plan of the 
upper curve of the elbow. 

For the patterns for the lower part of the elbow 
proceed as follows: Using as radii L 2, L 1 in plan and L 1 
in diagram W as center, describe the arcs 2 2° and 1 1° 
respectively, and draw a radial fine from 1 to the apex 



500 



The New Metal Worker Pattern Book. 



L\ cutting the inner arc at 2. Now take the various 
divisions along the compound curve 1 1' in the diagonal 
elevation, indicated from 1 to 11, and place these divisions 
on the outer curve in the pattern W as shown by similar 
numbers. From point 11 draw a radial line to the apex 
L 1 , intersecting the inner arc at 2°. Now take the dis- 
tance from 11 to the point of tangency 1" in the diagonal 
elevation, and place it as shown from 11 to 1" in the pat- 
tern W and draw a line from 1° to 2°. 1 2 2" 1" then 
gives the pattern for the side 1 2 in plan of the lower 
curve of the elbow. 

Using as radii M 2, M 3 in plan with M 1 in diagram 
X as center describe the arcs 2 2° and 3 n. Draw a line 
from 3 to the apex M 1 , cutting the inner arc at 2. Divide 
the inner arc 2 2" in the pattern W, as shown by the 
small figures 26 27, and transfer these divisions as shown 
by similar number on the inner arc 2 2" in the pattern X. 
From the apex M 1 draw a radial line through 2°, cutting 
the outer arc at n. Now measure the distance in the 
diagonal elevation from the radial line n to the point of 
tangency 3" and place it, as shown from n to 3° in X, and 
draw a line from 3 C to 2". Then 2 3 3" 2" will be the 
pattern for the lowc* ">art of the_elbow, indicated by the 
side 2 3 in plan. 

Using as radii N 3, N 4 and plan with N 1 in diagram 
Y as center, draw the arcs 3 n and 4 4" respectively. 
From 4 draw a radial line to the apex of N 1 , cutting the 
inner arc at 3. Now take the various divisions placed on 
the outer curve 3 n in X, and starting from 3 in Y place 
similar divisions on the inner curve as shown by similar 
letter and numbers. From the apex N 1 draw a radial 
line through n, intersecting the outer arc at 4°, and draw 
a line from 4° to 3°. The distance from n to 3" repre- 
sents the distance from n on the radial line in diagonal 
elevation to the point of tangency 3". 3 4 4" 3° then 
shows the pattern for the lower part of the elbow indi- 
cated in plan by the side 3 4. 

The final pattern for the curved elbow is obtained 
by using as radii 1 and 4 in plan and with O 1 in dia- 
gram Z as center, describe the arcs 1° 1 and 4° 4. Draw 
a radial line from 4 to O 1 , cutting the inner arc at 1. 
Now take the various divisions between 1 and 1° in pat- 
tern W and set them off on the inner arc in pattern Z as 
shown between 1 and 1°. From the center O 1 draw a 
radial line through 11, intersecting the outer arc at 4°, 
and draw a line from 4" to 1°. The distance 1" 11 equals 
the distance from the point of tangency 1° in diagonal 
elevation to the intersection between the radial line and 
curve at 11. 1" 1 4 4" is then the pattern for the lower 
part of the curve shown in plan by the side 1 4. The 



pattern Z may be further proven, if desired, by compar- 
ing the spaces between 4 and 4° on the outer curve in the 
pattern Y with those shown by similar numbers in pat- 
tern Z. 

The next step is to develop by triangulation the 
various sides of the straight portion of the elbow joining 
the curved parts. It will first be necessary to find the 
various true lengths of the lines shown in plan as follows: 
As the lines 1° 1° and 3" 3° in plan he on a horizontal 
plane, then will the corresponding lines 1" 1° and 3° 3° 
in diagonal elevation show their true lengths. Now, from 
the various points of tangency in the diagonal elevation 
as 1°2° 3° 4°, also 1* 2" 3" 4", draw lines to the left 
indefinitely parallel to C D, as shown. At pleasure erect 
any perpendicular as shown by P R. Now take the 
length of 2" 2° in plan and set it off on the line drawn 
from 2" in diagonal elevation, measuring from the line 
P R, as shown, from 2"° to 2", and draw a fine from 2" to 
where the line drawn from 2° in elevation intersects the 
line P R also at 2°. 2° 2° in the true lengths then shows 
the true length of 2° 2" in either plan or elevation. 

Proceed in similar manner for the balance of the lines 
in plan. Take the length of 4° -4° in plan and place it in 
the true lengths measuring from the line P R on the line 
drawn from 4° in elevation, as shown by 4' in the true 
lengths, and draw a line from 4" to where the line drawn 
from 4° in elevation intersects the line P R in the true 
lengths at 4°. This line from 4° to 4° is the desired true 
length. Now take the lengths of 2" 1° and 2" 3° in plan, 
and place them in the true lengths on the lines drawn 
from 1° and 3° in the diagonal elevati'on, measuring in 
each instance from the line RP, indicated in the true 
lengths by 1° and 3°" respectively. From 1° and 3° draw 
lines to where the line- drawn from 2° in elevation inter- 
sects the perpendicular P R at 2™, which gives the desired 
length. 

Finally take the length of the lines in plan shown by 
4° 1° and 4° 3° and place them in the diagram of true 
lengths, measuring from the fine R P on lines drawn 
from 1" and 3° in the elevation as shown by 1° and 3° in 
the true lengths. From these two points 1" and 3° draw 
lines to 4°, which point is the intersection between the 
line drawn from 4° in elevation and the perpendicular 
line R P. 

Having obtained all of the true lengths, the various 
patterns for each complete side is now in order, starting 
with the side 1' 2' at the top. Let the diagram S 1 be a 
reproduction of the pattern S. Now using 2° 2" in the 
true lengths as radius and 2° in S 1 as center draw the 
arc 2°, which intersect by an arc struck from 1° as center 



Pattern Problems. 



501 



and 1° 2 m in the true lengths as radius. Now, using 1° l r 
in the diagonal elevation (which shows its true length), 
as radius, and 1° in pattern S 1 as center, draw the arc 
1", which intersects by an arc struck from 2° as center and 
2° 1° in pattern W as radius. Take a reproduction of 
pattern W and place it, as indicated, by W 1 placing the 
line 2° 1" in W upon the line 2° 1' in W 1 . 1' 2' 2 1 is 
then the full pattern for that side. Take a tracing of the 
pattern T and place it, as shown, by T 1 . Using 2° 2" 
in the true lengths as radius and 2° in T 1 as center, de- 
scribe the arc 2\, which intersect by an arc struck from 3° 
as center and 3° 2" in the true lengths as radius. With a 
radius equal to 2" 3" in pattern X and 2° in X 1 as center 
describe the arc 3", which intersect by an arc struck 
from 3° in T 1 as* center and 3° 3° in the diagonal elevation 
as radius. 

Take a tracing of the pattern X and place it as shown 
by X 1 , placing the line 2" 3° in pattern X upon the line 
2" 3" in X 1 . 2 3 3' 2' is then the full pattern for that 
side shown by similar numbers in plan. In a similar 
manner take a tracing of the pattern U and place it as 
shown by U 1 , then using 3° 3" in the diagonal elevation 
(which shews its true length), as radius, and 3° in U 1 as 
center, describe the arc 3°, which intersect by an arc 



struck from 4° as center and 4° 3" in the true lengths as 
radius. With radius equal to 3" 4' in pattern Y and 3° 
in Y 1 as center draw the arc 4", which intersect by an arc 
struck from 4° as center and 4° 4" in the true lengths as 
radius. Take a tracing of the pattern Y and place it as 
shown by Y 1 , being careful to place the line 3"4*iriY 
upon the line 3° 4" in Y 1 . 3 4 4' 3' is then the pattern 
for similar numbered side in the plan. Finally take a 
tracing of the pattern V and place it, as indicated, by 
V 1 . Using 1° in V 1 as center, with 1° 1° in the diagonal 
elevation (which shows its true length), as radius, de- 
scribe the arc 1° in V 1 , which intersect by an arc struck 
from 4° as center and 4° 1" in the true lengths as radius. 
Now with radius equal to 4° 4" in the true lengths and 
4° in V 1 as center, describe the arc 4", which intersect 
by an arc struck from 1° as center and 1" 4" in pattern Z 
as radius. Take a tracing of 4' 1" 1 4 in pattern Z and 
place it as shown by similar numbers in Z 1 , being careful 
to place the line 1° 4" in Z on the line 1" 4" in Z 1 . 1 4 
4' 1' is then the pattern for the side shown by similar 
numbers in plan. Allow flanges on all patterns for rivet- 
ing, being careful to punch all holes accurately, as the 
material to be used is to be of number 16-gauge gal- 
vanized iron. 



PROBLEM 252. 
Patterns for a Branch Leaving in Two Directions. 



The tap is a 9-inch branch on a 30-inch main line 
pipe, and it leaves the center of the top of the main at 
an angle of 45 degrees and also points to the side of the 
main line at an angle of 45 degrees. It is a simple proposi- 
tion to solve in so far as the pattern cutting is concerned, 
because the parallel line system of developing surfaces 
of solids can be employed, once the view is obtained, in 
which the elements, always parallel in cylindrical objects, 
are shown in their true lengths. 

RSTU represents a plan view of the 30-inch joint 
of pipe of the main line, while W X Y represents the top 
center line of the joint. X Z is the axial line of the tap 
which is an imaginary line through the center of the tap. 
It does not touch the outer line or surface of the tap in 
any place, whereas W X Y is an element of the main 
joint and lies on the surface of the joint. 

P is the profile of the tap and is just placed there to 
show that line X Z is the tap. No other work need be 
done on this plan for obviously line X Z leans so that it is 



not seen in its true length and now the side elevation is 
projected from the plan and W' X' Y' is the top line of 
the main joint, while X' Z' is the axial line of the tap. 
Profile P' again is placed there just as an aid in recognizing 
the relation of the lines W' X' Z' and X' Z'. 

It is apparent when this view is studied, that line 
X' Z' leans both ways and is not shown in its exact length ; 
so, the pattern cannot be developed from this view and, 
therefore, the miter line need not be projected, as with the 
plan, unless a finished drawing is wanted. Even then 
this much of the plan and side elevation must be left as 
it is until the required view is obtained, when the various 
points of intersection and elements would be projected 
back from view to view. All that work is unnecessary, 
and as this is an attempt to show shop practice, where 
not more than one hour would be allowed to cut this pat- 
tern, the next step in the procedure will be described, 
although it is a good thing for those who have the time 
to construct all views complete. 



502 



The New Metal Worker Pattern Book. 



The next step, therefore, will be to project an end 
view or elevation. In this view, the profile PP, of the 
main joint is given and line W° X° Y° appears as one 
point only, while axial line X Z is shown as a line X° Z°. 
Once again, it is apparent that this axial line leans both 
ways and is not shown in the true length, and although 
this line is 45 degrees from the horizontal similarly to the 



the element, or line 3, is tangent to profile PP; had it 
occurred that element 3 fell outside of profile PP, it would 
have been necessary to shift line Z° X° until element 3 
would touch profile PP, which would mean that the rela- 
tive positions of line Z X would be altered in both the plan 
and side elevation. 

If now the axial line of the tap in the end elevation is 



/ 2 




Developing Patterns for a Tap. 



side view, it happened so because the line is 45 degrees 
from a relative line in plan. This would not be so if 
angle in the plan was 30 degrees, or any other angle for 
•then angle in end view would not be like either the one 
in side elevation or plan. 

Profile P° is now necessary and placed as shown and 
divided into equal parts. Lines parallel to Z° W° are 
drawn from these division points until they touch the 
profile PP of the main joint, all as shown. Note that 



viewed at right angles to it or along the direction of the 
arrow, it will be seen in its true length. Therefore an 
oblique elevation along the line of the arrow, projected 
from the end view, will be projected in this wise: Draw 
a line a b, parallel to Z° W°. Draw projector lines from 
points D and c of the profile PP, also from points Z° and 
W° of the end elevation. Make W x X x Y z of the oblique 
elevation equal to W X' Y' of the side elevation— that 
is to say, distance from W 1 to X 1 and then to Y x is the 



Pattern Problems. 



503 



same as W to X' and then to Y'. On the projector line 
from Z° in the end elevation and measuring from b in 
the oblique elevation, place the distance b' Z' of the side 
elevation, locating thereby point Z z in the oblique eleva- 
tion. Draw a line from Z 1 to X J . Complete the oblique 
view of the 30-inch joint, which will look and have the 
same dimensions as in the plan and side elevation. The 
heavy line representing the top line of the main joint 
and axial line of the tap having been projected from view 
to view, the miter line, or rather the line of intersection of 
the tap and main can be developed like this: 

Place profile P 1 in the oblique view as shown, with 
the same divisions as in the end elevation, being careful 
to number them correctly. Draw lines from these divi- 
sion points in profile P x indefinitely toward line Y x W. 
Draw projector lines from the points of intersection of 
the lines from profile P° with profile PP toward the 
oblique view, as shown, until they meet those drawn from 
profile P 1 , being careful to have like numbered projector 
line intersect that from profile P 1 . To illustrate, point 3 
on profile PP will intersect line from 3 in profile P 1 at 3* 
in the oblique view. Trace a fine through these points 
of intersection, as shown, which will give the desired miter 
line and correct view of main joint and tap with their 
elements shown in true length. 

The elements now being shown in true length the 
pattern can be developed by continuing the fine from 7/ 
and placing thereon the stretchout of profile P 1 , 1 2 3 4 5 
67891011 and 0. Draw parallel lines at right angles to 
the stretchout line from these points, which are inter- 
sected by the like numbered lines from the miter fine; 
like 3* is 3° in the pattern. A line drawn through these 
points of intersection completes the full pattern for the tap. 

Continue line b a and place thereon the half stretch- 
out of profile PP. Spaces ABCDE and MNO can 
be at pleasure, only care should be taken that these 
divisions are carefully placed on the stretchout line, 
especially those which are numbered. Draw lines from 



the numbered points on the stretchout line as shown, which 
intersect by lines from the miter fine — 3* is 3' in the pat- 
tern, and so on. A fine drawn through the points of inter- 
section will give the hole to be cut in the main joint. 

When shaping these patterns in the rolls, it is neces- 
sary to roll them so that they will have the right side out, 
otherwise the tap would point in the opposite direction. 
As the patterns now appear, the surface of the pattern of 
the tap showing should be outside when rolled, or in shop 
talk, "marks outside"; whereas, with the main joint 
the pattern surface now showing should be rolled so that 
that surface will be inside or " marks inside." 

The oblique view is complete, and to finish the others, . 
points on Z J are projected back to Z° in the end view, 
which would show the end of the tap, elliptical. These 
points in the elh'pse at Z° are then projected across to the 
side elevation to intersect points from P' which again 
would give a similar ellipse at Z'. Also the points on 
profile PP would be projected across to meet those lines 
from P' which would give a complete view of the tap in the 
side elevation. Likewise, all these points in the side 
elevation are projected down to the plan to either inter- 
sect fines from P or those swung around from the end 
elevation, which would then give a complete plan. As 
was said, this is unnecessary labor and would be dis- 
pensed with in shop practice. 

The procedure as explained here calls for a large 
drawing board which most well equipped shops have. 
Should, however, no large board be available, the work 
could be laid out on the floor, as is often done in many 
shops. On a small board the various views could be 
drawn singly, completing one view on separate sheets of 
paper and carrying the dimensions and the like from 
drawing to drawing. The actual pattern cutting could then 
be done right on the sheet metal, by scribing thereon the 
stretchout fines, and so on, and then carrying the lengths 
of the element lines by any convenient method. Or else 
the drawing would be quarter size and pattern full size. 



PROBLEM 253. 
Pattern for Irregular Transition Piece. 



The pattern here developed is for an irregular transi- 
tion piece which is to connect the opening in the blower 
to a flat pipe that runs along the ceiling of the basement, 
in a building where an indirect heating system is to be 
installed. The opening in the blower is an irregular 
.shape with five sides. The profile for one-half of the 



opening in the blower is shown in the cross section in 
Fig. 863. The shape of the main pipe which runs along 
the ceiling is also shown by cross section in Fig. 863. 
The plan of the blower and the transition piece is shown in 
Fig. 864. 

A close inspection and study of the proposed form of 



504 



The New Metal Worker Pattern Book. 



the transition piece as shown in the elevation in Fig. 864 
will show that if the corner of the outlet from the fan was 
connected directly to the far side of the pipe at the bottom, 
and also if the next corner, upon the outlet from the fan, 
was connected to the upper far corner of the pipe, the 
surface, which would form the side of the transition piece, 
would be vertical at the intersection of the flat pipe and 
on a slope at the intersection with the outlet from the 
blower. This would make the surface of that side of the 
transition piece a warped surface and only an approximate 
pattern could be developed, and this would necessitate 
the use of the triangulation method to develop the pattern. 







Pattern for an Irregular Transition Piece. 

It will also be easy to see that a direct line of con- 
nection between the two openings, as shown in the eleva- 
tion Fig. 863, would make an abrupt angle at the upper 
part of the connection and would decrease the size of the 
pipe so much as to interfere with its efficiency. There- 
fore, before a pattern can be developed, it is necessary to 
draw an arbitrary miter line. The most natural shape 
that suggests itself is formed by natural, graceful and 
gradual curves from the outlet in the fan to the end of the 
pipe. One miter line is drawn from X through the points 
ab c, etc., to I as shown in the elevation in Fig. 863, and 
the other one from through the points 2 3 4 and 5, etc., 
to 15. It will also be noticed that the fifth corner or high 
point of the outlet from the fan is not taken into consid- 



eration at this time. Next draw a miter line on graceful 
curves in the plan Fig. 864, which will represent the edge 
of the lower surface of the transition piece and is desig- 
nated in Fig. 864 by the figures 5, 6, etc., to 15. In order 
to keep the surface of the side of the transition piece from 
being a warped surface, the line which represents the edge 
of the upper part of the transition piece must be drawn 
exactly vertical and directly over the other line which 
represents the edge of the bottom of the transition piece. 
These lines which have been drawn arbitrarily may be 
taken for the miter lines of the transition piece and all 
that is necessary to complete the last run is to connect it 
with the point in Fig. 864. When 
these fines have been drawn the proc- 
ess of getting the pattern is very 
simple and consists of the following: 
Draw any fine A B at right angles with 
the parallel sides of the lines which are 
located on the sides of the transition 
piece. The stretchout for the far 
sides of the transition piece should 
now be laid off on this fine A B, the 
spaces being taken from the plan in 
Fig. 864 on the line 0', 2', 3', 4', etc., 
to 15'. The first space will be from 
0' to 2', the second from 2' to 3', the 
third from 3' to 4', etc. 

It will be noted that the spaces 
on the fines in the plan in Fig. 864 
which occur on a short curve, have 
been placed nearer together. These 
spaces do not need to be equal when 
it is convenient to make the work 
easier. It will also be noted from the 
point to the point 2 in the elevation 
in Fig. 863, there is no division and this 
is explained by the simple fact that when a fine is straight 
in the plan and also in the elevation, it will be straight on 
the pattern. The points on the sharper parts of the curve 
are located closer together to make the pattern more 
accurate. Now, from the points in the plan Fig. 864 on 
the line from 0', 2', 3', 4', etc., to 15', project lines down 
vertically until they intersect with the edges of the transi- 
tion piece in the elevation Fig. 863. Draw lines through 
these points parallel to the fine A B in Fig. 864, until 
they intersect with lines from the stretchout for pat- 
tern. A line through the points thus obtained will be 
the true shape of the pattern, as shown in Fig. 865 for 
the larger side of the transition connection. 

The stretchout for the bottom of the transition con- 



Fig. S65. 
Pattern for Larger Sid 



Pattern Problems. 



505 



nection is taken from the elevation in Fig. 863 and laid 
off on the line C D which must be drawn parallel to the 
line A B and at right angles to the end of the flat pipe and 
to the lines which have been drawn on the side of the 
transition piece in the plan and elevation as shown. The 
stretchout for X a b c d, etc., to I is taken from Fig. 863 
and copied onto the line CD in Fig. 866. Lines drawn 
vertically upward from the points on the side of transi- 
tion piece, as shown in Figs. 863 and 864, are projected 
over to the pattern parallel to the line C D until they inter- 
sect at the stretchout line of the pattern. The line 
traced through these points will be the true shape of the 
pattern for the bottom of the transition piece as shown in 
Fig. 866. The stretchout of the pattern for the top of the 
transition piece, as shown in Fig. 867, is taken from the 
true profile of the top, as shown in the elevation in Fig. 863 
in the same manner as in the preceding pattern. The 
location of the points on this pattern are determined, as 
in the preceding case, by projecting lines from the points 



in the plan Fig. 864 to the pattern in Fig. 866. It will be 
noted in Fig. 867 that a cut has been made in the pattern 
for a part of the transition piece over which a " boss " 
is to be placed that will connect the fifth point of the 
outlet from the fan to the transition piece. The pattern 
for the boss is shown in Fig. 868. The stretchout for the 
pattern in Fig. 868 is taken from the section through the 
outlet, one-half of which is shown in Fig. 863 by cross- 
hatched section. The points on the pattern for the boss 
in Fig. 868 are located by projecting them from the ele- 
vation of the boss in Fig. 863. The projection must be at 
an angle of 90 degrees with the hne which runs from point 
1 to point 2 in the elevation Fig. 863. The line E F, 
which is to be the center of the pattern, must be parallel 
to the line that runs from 1 to 2 in the elevation Fig. 863. 
No allowance has been made in the pattern for any of the 
parts for the transition piece to be secured together and 
the necessary laps must be added by the mechanic to suit 
the requirements of the method of fastening which he uses. 



PROBLEM 254. 



Pattern for Forty-five Degree Y Branches. 



This problem has to do with branches leaving a main 
pipe at an angle of 45 degrees. In the plan A shows the 
profile of the main, 10 inches in diameter, and W Z B C 
shows it in side elevation, from which two 5-inch branch 
pipes, E and D, are taken at an angle of 45 degrees. The 
distance that the bottom of the pipe E sets below D is 
indicated by X. The branch pipes incline at an angle 
of 45 degrees in both plan and elevation, as indicated 
by E 1 and D 1 . They are so placed that a hne through 
the center of branch E 1 will meet the side of branch D 1 
extended at center A, so that the distance from the center 
hne at Y to the side of pipe E 1 equals Y 6', while the 
distance to the side of pipe D 1 will equal Y 1'. 

As none of the views in either plan or elevation shows 
the true lengths Or true angles of the branches, a true 
elevation must be found showing the true length and 
true angle between the branch and main, from which the 
patterns are obtained, as follows : 

Through the center of pipe D 1 in plan, draw the 
center hne 3' 3, on which establish H as center and draw 
the required size circle. In a similar manner draw the 
center line through branch D in elevation, as shown by 
3 1 a, and using a as center draw the circle shown. Estab- 
lish any point as F on the center line 3' 3 in plan, and 



from this point and the intersection between the center 
line and main pipe at 3', erect vertical lines intersecting 
the center line in branch D in elevation at F 1 and 3 1 
respectively. From Z x draw a horizontal hne, cutting the 
vertical line, previously drawn, at G and forming the 
right angle triangle F 1 G 3*. With points 3' and F in 
plan and G and F 1 in elevation located nothing further 
is required in finding the true length and angle. 

Equal in length and parallel to 3' F in plan draw the 
hne 3* F 2 in the true elevation. At right angles to this 
line from F 2 erect the perpendicular F 2 F 3 equal in night 
to G F 1 in elevation. Draw a hne from F 3 to 3* in the 
true elevation, which on its center hne will give the proper 
or true pitch of the branch pipe shown by D in elevation. 

The true joint or intersecting hne between the branch 
and main pipe is found as follows: Extend the center line 
in the true elevation as 3 X c and using c as a center draw 
the circle H l , a duplicate of H in plan. Divide both of 
the circles H and H 1 into an equal number of spaces, as 
from 1 to 5 to 1 being careful to have the numbers so 
placed that if 3 and 3 come on the center hne in profile 
H, similar numbers 3 and 3 will be at right angles to the 
center hne in the profile H 1 . From the intersections 1 to 
5 to 1 in the profile H draw fines parallel to the center line 



506 



The New Metal Worker Pattern Book. 



3 3' until they intersect the profile of the main pipe from 
1' to 5' and from which points lines are projected to the 
true elevation at right angles to 3' 3 in plan, and inter- 
sected by lines drawn from similar numbers in the profile 
H 1 in the true elevation parallel to the center line c Z x , 
resulting in the points of intersections 1, 2, 3, 4, and 5 on 
both sides in the true elevation. A line traced through 



Openin 
large Pi 



Opening in 
large Pipe 




Pattern for Main Pipe in one Piece 




Fig. 869. — Patterns for Y Branches Leaving Main at Different Points. 

points thus obtained will give the joint or miter line, from 
which the patterns are obtained. 

For the pattern for the branch pipe proceed as fol- 
lows: At right angles to 5 F 3 in the true elevation draw 
the line J K, on which place the stretchout of the profile 
H 1 , as shown by similar numbers. At right angles to J K 
from the various intersections 1 to 5 to 1, draw lines which 
intersect by lines drawn parallel to J K from similar inter- 
sections on the joint line in the true elevation. A line 



traced through these points, as shown by K L M N J, 
will be the pattern for the branch pipe. 

The opening in the main pipe is obtained as follows: 
At right angles to the line of the main pipe in the true 
elevation draw any line as P, on which place the girth 
of that portion of the main pipe in plan shown from 1' to 
5', measuring each space separately, as they are all unequal. 
At right angles to P draw lines from the small figures 
which intersect by lines drawn parallel to P from 
similar numbers on the joint line in the true elevation. 
The shaded portion indicates the pattern for the opening 
which is traced through points thus obtained. 

Establish any point on the center line of pipe E 1 
in plan as d, which use as a center and describe the 5-inch 
diameter circle, as shown. In 
a similar manner chaw the center 
line through pipe E in eleva- 
tion, as shown by 8 1 b and using 
b as center, draw the circle 
shown. Establish any point as 
S on the center line of branch 
E 1 in plan and from this point 
and the intersection between the 
center line and the profile of the 
main pipe at 8' erect perpen- 
dicular lines, intersecting the 
center line of the branch E in 
elevation at S 1 and 8 1 respect- 
ively. From 8 X draw a horizon- 
tal line, intersecting the per- 
pendicular line previously drawn, 
at R, and forming the right triangle S^S*. 

The true elevation may now be found as follows: 
Equal in length and parallel to S 8 1 in plan, draw the 
line S 2 8 X . At right angles to S 2 8* from the point S 2 
set off the distance S 2 S 3 , equal in hight to R S 1 in the 
side elevation. Draw a line from S 3 to 8 X in the true 
elevation, which will give the true pitch on its center 
line of the branch pipe shown by E in elevation or E 1 
in plan. 

The true joint or miter line between branch E and 
main pipe is found in the true elevation, as follows: 
Extend the center line in the true elevation 8 X d' and 
using d' as center draw the circle, which is a duplicate 
of d in plan. Divide both of the circles d and d' into 
an equal number of spaces, being careful to have the 
numbers so placed that if 8 and 8 come on the center 
line in d similar numbers 8 and 8 will be at right angles 
to the center line in the profile d'. From the inter- 
sections 6 to 10 in the profile d draw lines parallel to 



Pattern for Branch Pipe, 



Pattern Problems. 



507 



the center line 8 8' until they intersect the profile of 
the main pipe from 6' to 10', as shown, and from these 
intersections lines are drawn at right angle to 8 8' into 
the true elevation, and intersected by lines drawn from 
similar numbers in the profile d parallel to the center 
line d' 8. A line traced through points thus obtained, 
as shown, from 8 to 6, 10 to 8 will give the joint line 
between the branch and main line. 

To obtain the opening in the main pipe for the 
branch proceed as follows: Take the girth of the various 
spaces between 6' and 10' in the plan view of the main 
pipe and place it on the line h i at right angles to the 
lines of the main pipe in the true elevation. At right 
angles to h i, from the small figures, 6' to 10', draw lines 
which intersect by lines drawn parallel to h i from sim- 
ilar numbered intersections in the joint line in the true 
elevation. Trace a line through these points, as shown 
shaded, which will be the desired opening. 

The pattern for the branch is obtained by drawing 
any line at right angles to S 3 8 X , as shown, by T U and 
on this the girth of the profile d! is placed. The usual 
measuring lines are drawn and intersected by lines drawn 
parallel to TU from similar numbered intersections 
on the joint line, 8 8, in the true elevation. The pattern 
shape is represented by TU/j. 



As the two branches intersect the main pipe in 
different positions, namely, one tangent and the other 
central, with one branch higher than the other, it will 
be necessary to lay out the pattern for the main pipe 
in one piece, with the openings placed in their proper 
positions. 

Take the hights from W to 3° to Z in the side 
elevation and place it as shown, by similar numbers in 
pattern A. Take the hight of X in elevation and 
place it, as shown, from 3° to 8° in (A). At right angles 
to Z W from 3° draw a line indefinitely to the right, 
while from 8° draw a line indefinitely to the left. Take 
the girth from Y to 1' and Y to 6' in plan and place 
it, as shown, from 3° to 1' and 8° to 6' respectively in 
(A). From 1' and 6' erect perpendiculars, as shown. 
Take a tracing of the opening B and place the angle 
1' 1' 3° on similar angle 1" 1' 3' in the pattern A. In 
a similar manner take a tracing of the opening C and 
place the angle 8° 6' 6° on the angle 8" 6' " in the pattern 
A. Take the girth of the semi-circle Y to v in plan 
and place it on either side of the center fine Z W in 
A, as shown, by V 1 V 2 and W 1 W 2 , which completes the 
pattern for the vertical pipe in one piece. 

A lap must be added to all patterns for seaming 
and riveting. 



PROBLEM 255. 
Pattern for Offset Blower Connection. 



In a large building, the existing conditions made it 
necessary to reduce the size of a blower pipe in a given 
distance and at the same time to drop it down from the 
ceiling and to shift it out from the wall with the result 
that the finished connection appeared as shown in Fig. 870. 

The method of developing the patterns for the re- 
quired offset and reducing section is as follows: 

Draw the plan and elevation as shown in Fig. 871 
and divide the line X Y into any number of equal parts. 
Extend these lines up and down, as indicated by arrow, 
until they intersect the plan and elevation on each side 
as shown. To develop the pattern for the top, take the 
distances 1, 1 2, 2 3, 3 4, 4 5, 5 6 from the profile of 
the top in the elevation and lay them off on the fine A B 
located at the right of the plan. Care must be taken to 
transfer them in the proper order, as the spaces are now 
unequal. Carry the lines over from the points 0', 1', 2', 
3', 4', 5', 6', in the plan until they intersect with lines 



drawn at a right angle from the points located on the true 
stretchout line A B. 

By tracing a line through the intersection of these 




Fig. 870. — Finished Blower Connection. 

lines, the correct outline is developed for one side of the 
pattern for the top of the connection. In like manner, if 
lines are carried over from the points A', B', C, D\ 



508 



The New Metal Worker Pattern Booh. 



E', F', G', in the plan, a line traced through the points of 
intersection of these lines, and lines extended from the 
points 0, 1, 2, 3, 4, 5, C, on the line A B, the resulting 
figure will be the correct outline of the other side of the 



the bottom of the connection as shown on the hne C D 
must be taken from the profile of the bottom as shown 
in the elevation. The spaces on the pattern for the front, 
as shown on line E F, must be taken from the profile of 




F T 6' 
Fig. 871.— Plan, Elevation, and Patterns for Offset Blower Connection. 



pattern for the top of the connection. Now by adding 
the proper amount of material from the points and 6 on 
the line A B to make the straight sections on either end 
of the connection, the pattern is complete. 

To develop the patterns for the other parts, simply 
repeat this operation. The spaces on the pattern for 



the front, as shown in the plan. Also the spaces for the 
pattern of the back as shown on the hne G H must be 
taken from the profile of the back as shown in the plan. 
No laps are added to the patterns in this solution as the 
practice of different shops for fastening corners together 
calls for different allowances. 



PROBLEM 256. 
Pattern of a Raking Soffit Bracket. 



A finished view of a right and left raking soffit 
bracket is shown in Fig. 872, such as is used in the soffit 
of bay windows, as shown in plan in Fig. 873. In this the 
soffit plan of part of the base mold is shown, as well as 
the normal bracket in the center, the right and left raking 
brackets C and D, and the panels. A sectional view 
through A B in plan is shown by a b c in the section and 
represents the given profile of the center or normal bracket, 
from which the various measurements and joints of 
intersections are obtained in developing the raking 
bracket. 

The various steps required in developing the patterns 
for the raking soffit bracket are shown in detail in Fig. 874, 



in which A, 2, 26, shows the true profile of the normal 
or center bracket shown by A B in plan in Fig. 873. 
Now, divide this normal profile in Fig. 874 in equal 





Fig. 872.— View of Bight and Left-hand Soffit Bracket. 



Pattern Problems. 



509 



spaces as shown by the small figures 1 to 27 and in line 
with this normal bracket place the plan view of the raking 
bracket, as shown by C, B, 1, 27, which represents the 
raking bracket shown by D in the soffit plan in Fig. 873. 
Now, from the small figures 1 to 27 in the normal bracket 
in Fig. 874 erect perpendicular lines until they intersect 
the sides of the bracket 1, 27 and C B in plan, as shown 
numbered on the lower side. At right angles to the lines 



dividers equal to the width of the face 1 B in plan and step 
off this distance on every one of the lines, 1 to 27, in the 
pattern. This makes the cut B H similar to F G and 
F G H B becomes the pattern for the face of the raking 
bracket, which can be formed right and left as desired. 

The pattern for the side of the raking bracket is 
now in order and is obtained as follows: Take the length 
of the line 1 27 in plan with the various intersections on 



,12 3456 7S9IO 



J 

*4, 



: 






Soffit 
Plan 




Sectional View 
through A- B 




Pattern for Sides 
of Raking Bracket 



Fig. 873. — Soffit Plan and Sectional 
View of Raking Brackets. 



27 hx 

26 25 

Fig. 874. 



-Developing Patterns for Side and Face of Raking Bracket — Note Plan and Profile 
of Normal Bracket — Finished View is Shown in Fig. 872. 



just drawn, draw the stretchout line D E, upon which 
place the girth of the normal bracket 1 to 27, as shown by 
similar numbers on D E. From these intersections, at 
right angles to D E, draw fines which intersect by lines 
drawn parallel to D E from similar numbered intersec- 
tions on the side 1 27 in plan. Trace a fine through points 
thus obtained; then F G will be the cut of one side of the 
face of the bracket. 

The opposite cut B H could be obtained from the 
various intersections on the side C B in plan, but in this 
case the cut B H has been obtained as follows: Set the 



same, and place it at right angles to the line A 26 in the 
normal profile, as shown by the similar intersections on 
the line J K. At right angles to J K, from the various 
intersections on same, drop fines, which intersect by lines 
drawn parallel to J K from similar intersections in the 
normal bracket. Trace a fine through points thus 
obtained ; then L M N will be the pattern for the sides 
of the raking bracket. The pattern for the front of the 
raking bracket is formed after the normal profile, and 
the pattern for the sides are soldered along the miter 
cuts in the face pattern, as shown by a in Fig. 872. 



510 



The New Metal Worker Pattern Book. 



PROBLEM 257. 



Patterns for Offset Y from Circular Pipe. 




Side 
Elevation 



Top Piece ■ 



One of the users of a previous edition of The New 
Metal Worker Pattern Book asked for method of de- 
veloping patterns for a Y from a round pipe which 
offsets from the main in two directions as indicated in the 
sketch reproduced in Fig. 875. He was unable to locate 
the problem in which the principles 
underlying his problem were given. As 
others may have the same difficulty this 
new problem has been included. 

As has often been stated in this 
book and many other publications, pat- 
tern drafting is the manipulation of the 
principles of geometry, and the most 
difficult part of the procedure in develop- 
ing the patterns for any object is the 
obtaining of view, or views of it from 
which certain and exact measurements 
may be taken. In other words, it is the 
determining of the true lengths of the 
elements embraced in the surface of the 
object, for pattern drafting is simply 
laying the surface out flat. Many sub- 
mitted problems are facsimiles of prob- 
lems in the book. Just because a profile 
may be differently situated, or the offset 



the true lengths of the elements. In Pig. 876 is shown 
the final study of the object. In this figure there is a 
front elevation which obviously does not give the true 
lengths of the elements of the middle piece, for these 
elements lean away from the observer. Nor does the 



Profile Profile 

Fig. 875. — Correspondent's Sketch. 




Oblique Elevation 
Fig. 876. — Development of Several Views of Offsetting Y Branch. 



is in another direction, the owner of the book fails to rec- 
ognize the similarity. 

A problem similar to the one under discussion is 
Problem 45 herein. However, that Y branch does not 
offset, consequently in the problem submitted a some- 
what different procedure must be followed, although the 
patterns would be developed by the same parallel line 
system once the view of the object is obtained showing 



plan give these lengths, or again, neither does the side 
elevation. Naturally, the question arises, what is neces- 
sary to obtain these lengths? 

In projecting various views of cylindrical objects, 
it is best to work from one element of the cylinder, prefer- 
ably the center or axial line and, after projecting this 
axial line to build about it the cylinder proper. In the 
various views of Fig. 876 the axial line is represented by 



Pattern Problems. 



511 



the heavy three dot and long dash line, the ends of these 
lines and their intersecting points are represented by 
heavy dots surrounded by a small circle and attention 
is called to how these axial lines appear in the various 
views and also the relative positions of the cylinders about 




Fig. 817. — Development of Patterns. 



these axial lines. When a line is viewed so that the line 
of vision, which is an imaginary line from the eye of the 
observer, is perpendicular or at a right angle to the line, 
that line is viewed in its true length regardless of the 
position of the line. Then, if the axial line of one of the' 
middle pieces of the object is viewed at right angles to its 
axial line in plan, as shown by Arrow A, that line is seen in 
its true length and an elevation projected in the line of the 



arrow will have the axial lines of that piece in true length. 
It would be the axial line of the one middle piece but not 
of the other. It would also give the axial lines of the 
top piece and the two bottom pieces because they he in 
parallel planes to the plane in which the axial line of the 
middle piece in question lies. Such a view is shown in 
Fig. 876 and is called an obhque elevation and is what 
is required. 

A drawing is given which is like those made by the 
cutters in the big shops who must economize in tune and 
yet make drawings that can be understood. This is 
given in Fig. 877 with the accompanying description. 
Before leaving Fig. 876, however, those interested in pro- 
jection should know that the drawing reproduced in Fig. 
876 must be clear enough to follow the various steps, the 
oblique elevation being commenced by first projecting 
the axial lines. Note that after projecting the points 

in plan B B B, points on these 
projectors are chosen at will as 
B° and B°° both on a line at 
right angles to the projectors as 
shown by the heavy- dotted 
fine. The hight then from 
this line to B°° is the hight 
given in the front elevation from 
the arrow point D, up to D°. 
Now then, for the practical 
treatment of the problem and referring to Fig. 877, it 
will be assumed that the line 2 K is the center or 
axial of the top piece of the branch, and that angle 
X is the approximate branching off of the middle pieces 
and Y T the axial line of the bottom piece, Y K of 
course, being the axial line of the middle piece, giving 
then K W° as the vertical hight of the points of 
intersection K and Y or the axial lines. Point 3' in the 
plan represents the point of intersection K and also the 
axial line in plan of the top piece. Point S in the plan 
is the center of the axial fine of the bottom piece at point 
of intersection Y. It must be understood that the middle 
and bottom piece of the other part of the object need not 
be represented as the idea is to do only such work as is 
absolutely necessary. At the right of plan R shows the 
amount of offset of the branch. About S draw a circle 
the size of the pipe which then depicts the plan view of the 
bottom piece. Do likewise about point 3' one-half circle 
only being necessary which then is a plan of one-half the 
top piece. Divide these profiles into equal spaces as 
shown, and draw the sides of the middle pieces and also 
its elements. Continue the elements 2, 3, and 4, to the 
center line. This center line represents the edge of a 



> Full Pattern 

of Top Piece 

/ required 



512 



The New Metal Worker Pattern Book. 



vertical plane against which the elements terminate as 
V 2' 3' 4' 5' and 6'. Parallel to the axial line in plan 3' 
S, draw a line as W 12°. Project a line at right angles to 
this and from 3' passing through W and of indefinite 
length. Do the same from S stopping at 12°. Set off 
the space W 12" equal to W° K of the front elevation. 
Connect 12° 12" in the oblique elevation which then is the 
required view of the axial line of the middle piece and 
shows the true length of the axial line. The space 6" 
M' and 6° M° in the oblique elevation coincide with K 2 
and Y T of the front elevation. This now gives a true 
view of the axial lines and consequently, the true angle 
they form with each other. Bisect these angles in the 
customary manner, that is, place the compass on 12° 
and swing arcs N' and N°. Extend the compass a bit 
farther, place the point on N° and describe a small arc as 
N which is bisected by an arc described with the com- 
passes at the same adjustment and the point at N'. 
Draw a fine from N through 12° which then is the miter 
line between the bottom piece and the middle piece. Do 
the same for the angle of the top and middle pieces giving 
miter line 1" 9". Project the element points 1 to 12 
to the miter line 3° 9° and thence parallel to the axial 
line to miter fine 1" 9". Project 1' in plan to inter- 
sect 1" in the oblique elevation 2' to 2", 3' to 3", 4' 
to 4" and so on, always being careful to intersect the 
right lines realizing the complete oblique elevation of 



one bottom piece, one middle piece and one-half of the 
top piece. 

For the patterns draw any line as a to a at right angles 
to the axial line in the oblique elevation of the middle 
piece. On tins line set the spaces of the profile as 1 to 1 
on fine a a. Draw the usual parallel lines through these 
points which are intersected by lines projected from the 
oblique elevation as shown, always being careful to 
intersect the right fine. 

Now the pattern so obtained for the middle piece will 
also do for the top and bottom pieces. The bottom cut 
is for the bottom piece as indicated by the note on the 
drawing. The part from A to B of the top cut is one-half 
the pattern for the top piece, two of this part joined make 
the whole pattern. To better show what is meant, how- 
ever, the top piece pattern is developed to the right of the 
oblique elevation though it is not necessary and in prac- 
tice would not be done because all the patterns are con- 
tained in that of the middle piece. 

The plan of the other middls piece of the branch could 
have been used to project the oblique elevation, but the 
same elevation would have resulted. Similarly, the 
oblique elevation could have been projected to the other 
side of the plan, reverse of the arrow A in Fig. 876, but 
again the result would be the same, for the method always 
comes back to the principle of projecting the view with the 
line of vision at right angles to the axial fine in plan. 



PROBLEM 258. 



Pattern for Range Hood with Reduced Sides. 



The front and side elevations and plan presented 
in Fig. 1 show the dimensions for the hood to go over a 
range. The hood is to be 34X45 inches and 27 inches 
high. The front is a regular circle, but it is desired to 
carry the sides in on the back at the top but 7 inches, 
so that the side curve will be as shown in the eleva- 
tion for the side at the back. This will make quite a 
different curve, and the problem is to cut out the pattern 
for this end as well as the front piece. 

The problem is similar to that of developing a pat- 
tern for a molding having a square return with less pro- 
jection thai: the front. In this case the front molding is 
composed of but one member, the profile or shape of 
which is an arc, the hight of which is 27 inches, while 
the projection or width is to be 34 inches; the length, of 
course, of the hood is not a governing factor, and the 



return of this molding is to be of a shape similar to the 
front, but is to have but 7 inches projection, with the same 
hight as the front. 

Now, such problems can be solved in two ways, 
although, in either method the patterns are developed by 
the parallel-line system. One way is by deciding on the 
shapes of both front and side and projecting parallel 
elements in each profile and at right angles to each other, 
to meet in a plan the result of which will be the producing 
of a required miter line, which, however, will be an irregu- 
lar fine, as shown. 

As may be supposed, an irregular joint line be- 
tween the side and the front is going to give consider- 
able trouble in making the seam, and the only advantage 
will be in having a profile for both front and side as pre- 
determined. The other way would be to draw a plan, as 



Pattern Problems. 



513 



in Fig. 878, which has a straight line for the miter or joint 
between the front and side, which, obviously, is highly 
desirable. However, but one profile can now be as 
chosen, while the other is modified and controlled by a 
developing process from the normal or chosen profile. 

It would seem that it really does not matter just what 
the contour of the sides are so long as they have the 
specified projection of 7 inches. Therefore, in Fig. 879, 
A B C D is the outline of the plan of hood drawn to scale, 
viz. : A to D is 34 inches and A to B 45 inches and A to 1" 




34'^ T 

Elevations 



Fig. 878. — Problem. Fig. S79. — Pattern for a Symmetrical Hood. 

is 7 inches; also B to 1°° is 7 inches. The miter lines 
drawn from 1°° to C, and 1" to D complete the plan. 

Deciding that the profile of the front shall be arbi- 
trarily established, rather than that of the side, let 1' 
7 be a base line of profile XX and 1' to 1 the hight of the 
hood, which is 27 inches. It is required that this profile 
be part of a circle, and as the hight is less than the pro- 
jection the profile cannot be a quadrant, but less than a 
quarter circle. So then set the compasses, or trammels 
in the actual work, to more than 34 inches, and, with the 
point at 7, strike a small arc outside of the diagram. 
Setting the point of the compass at 1, and keeping the 



same radius, intersect the arc just described, getting 
thereby point a. With a as center, and keeping the 
compass set to the same radius, describe the arc from 1 to 
7', completing the profile of the front. In range-hood 
designs, or, in fact, drawing profiles like these which have 
much less hight to projection, or vice versa, it is cus- 
tomary to set the compass to the smallest dimension, as 
27 inches. Then 27 inches from 1 is at b, and from b a 
vertical line is drawn ; also from b a quarter circle of 27-inch 
radius is described from 1 to c, as shown dotted. This 
gives a flat deck from c to 1, which is a good thing in hood 
designs; and again, when the side profile would be devel- 
oped, it will be found that that, too, will have a more 
pleasing contour, not changing so abruptly at 1° in the 
full profile on line A B. That, however, is a matter of 
taste, so to speak, for it does not matter what shape the 
front has in this procedure. It is to be 
noted that in drawing the arc 1 to 7 a 
radius greater than 34 inches was pre- 
scribed, because, if 34 inches — point d then 
as center — was used for the radius, the arc 
would dip down somewhere between points 
2 and 1, which certainly would not be a 
good feature in hood design. 

Divide profile XX into equal parts, as 
shown by 1 to 7, project these points to the 
left and through the plan. Mark with points, as 
shown, where these lines cross the base line of profile 
XX and where they intersect the miter lines. From 
where they intersect the miter line D or 7", to 1" of the 
plan, and at right angles to line A D, project the points 
upward indefinitely. Draw the line 7° B 1 , and then 
take the lengths from the base lines to the arc in profile 
XX and place them on like numbered lines in profile 
X, measuring from line 7° B', like this: 1' to 1 in profile 
XX equals in length l x to 1° in profile X; again, 2' to 
2, in profile XX, equals in length from 2 X to 2°' in 
profile X, and so on. A line traced through points 7° 6° 
5° 4° 3° 2° and 1° will be the modified profile of 7 inches 
projection. Complete the section, or profile X, as shown. 
For the patterns proceed this way: Extend line B C 
downward, and place thereon the spaces from the arc of 
profile XX, as l A to 7"\ Parallel to line D C draw lines 
from the points in the stretchout line. These parallel 
lines are intersected by lines projected from the miter 
line in plan, like line from point 3" in the plan is to inter- 
sect the line from point 3 A in the stretchout line. Lines 
through these points complete the pattern. 

Extend line A B to the left and place thereon the 
spaces 7° to 1° in the profile X, taking each separately, 



514 



The New Metal Worker Pattern Book. 



as they all differ from each other, as shown from 7 B to 1 B . 
Draw lines parallel to line A D from these points and 
intersect like numbered lines from points in the miter line 
in plan, like line from point 3" in the plan is projected to 
line from point 3 B in the stretchout line. Trace a line 
through the points and obtain the pattern as shown. To 



prove that the pattern is correct, measure the miter cut 
of each pattern, and if they coincide in length both pat- 
terns are correct. Pattern for ends would be shaped 
on line 7 s I s to the profile XX at 1° 7°, and pattern for 
front would be shaped on line H K to the profile XX, 
from 1 to 7. 



PROBLEM 259. 
Marquise Construction and Patterns. 



/There is no end to the number of designs possible for 
marquise or entrance canopies and the problem is essen- 



ever, and, therefore, requires attention and discussion. 
The design presented here is a very common one and, of 




\ 



Sheet Metal Molding '^Olass pendants in stamped zinc -frame 

Fig. 880. — Front Elevation of Marquise and Glass Work in Arch. 




Fig. 8S1. — Side Elevation of Marquise. 



r. Ouilet_ 



'y+mre 



Glass 



on 2. 



2'Ts 



Fig. 882. — Architect's Scale Drawing of the Marquise. 



tially one of construction rather than pattern cutting. 
It is an important object in the sheet-metal trade, how- 



course, can be elaborated more or less, as required. It is 
reproduced as actually constructed. 



Pattern, Problems. 



515 



The architect's sketches are reproduced in Figs. 
880 to 882, which show the front elevation, side eleva- 
tion and plan and give a good idea of the general design. 
Figs. 8S3 to 886 are groups of diagrams each being a 
detail section through some important part of the 
marquise. 

The patterns for the bars in the arch, all caps, 



this profile, draw a vertical line to represent the wall. 
Then, from the numbered points in the front profile 
draw parallel lines to the wall line. These parallel lines 
are to be pitched to the amount of inclination shown in 
the side elevation of Fig. 881. Note that from such hori- 
zontal members, like 7 8 or 13 14, two lines result instead 
of one, as in ordinary square return problems; this is due 





a 



Glasi 



Paint Skin Joint 
Sheet Mete! Flashing 



Sheet 
Metal- 



'■A||r||A / 'Wood Nailing 
,, Soffit sfn P 




Glass .■••"■ 
Pendants \ 



Upper Light Loner Light 

of Glass \ Cl/p,& : of Glass 



Mil 





Fig. 883. — -Vertical Section through Longest 
or Front Section of Molding. 



Fig. 884- — Vertical Section through the 
Raked Side Moldings. 



Fig. 885.— Vertical Cross Section at 
Central Channel Iron. 



cross clips, gutter strips and so on, are merely the 
stretchout of these sections given in the diagrams, be- 
cause no miter cuts, or cuts that are very simple, are 
required. 

The patterns of the front and side moldings are given 
in Fig. 887. The first thing to do is to draw the profile 
of the front elevation, as shown. This profile is divided 
into spaces and numbered. A convenient distance from 



to the raking of the side molding, which perhaps is 
better shown in the problems herein of raking pediments, 
only in those problems the pitch is the opposite to this 
so that the square members of the profile in those cases 
would be more than square in the raked profile, while in 
this case they are less than square in the raked profile, as 
may be seen in Fig. 887. 

The miter cuts on both ends of the front molding are 



516 



The New Metal Worker Pattern Book. 



square miters like Problem 3. Therefore, draw a stretch- 
out line, like A B of Fig. 887, which should be vertical, that 
is parallel with the vertical members of the front profile, 
as 16 17 or 23 24. On this stretchout line place the 
spacings 1 to 28 of the front profile, as 1° to 28°. From 



Skylight 
Bar 






1 

t 


V 


.-'' 




y\ 








Bolt 


.■" < 




y 




\ G/3! 


S-* 


• 







Vertical Section of 
Work at Arch Soffit. 



Horizontal Section of Bar in Arch 




Fig. 881. — Vertical Section showing Contour of Gutter, Arch Work 
and Other Details. 

Additional Details of Construction. 



the points 1° to 28° draw horizontal lines which intersect 
by projector lines from like numbered points in the front 
profile. For instance, point 10 in the front profile is 
projected down to horizontal line from point 10° on the 
stretchout fine, giving point lCf, and so on. A fine drawn 
through these intersection points completes the pattern 
cut of the ends of the front molding. 




Fig. 887. — Procedure for Developing the Patterns of the Molding. 



The New Metal Worker Pattern Book. 



51< 



A horizontal line is drawn under the profile, as 
shown by C D and where the projector lines, from the 
front profile, cross this fine, small dots are made to indi- 
cate the intersections. This fine, C D, is then trans- 
ferred over the rake fines as shown by C x D 1 , being sure 
that it is parallel with the rake fines. Of course, the 
front profile could be transferred to this position, as 
directed in the problems herein of raking pediments, but 
this method is more expeditious. It is, however, neces- 
sary to be very careful with this method so that the 
proper points are projected down from line C x D x . For 
that reason it is probably best to number the points on 
both fine C D and C z D x . From points in line C x B x 
drop fines, at right angles to C x D x , to intersect like 
numbered lines from tfie front profile. Draw a line 
through these points which will produce the raked profile 
of the side moldings, as shown. 

The soffit part, 24 to 28, is not raked owing to the 
absolute need of having the soffits of the side moldings 
level to fit the structural work to which they are fastened. 
The slight misfit at the miters, caused by this, would not 
be of any consequence because the pendants and electric 
lights cover most of the soffits. 

For the pattern of the sides, draw stretchout line, 
F G, at right angles to the rake or pitch lines of the side. 



Place on this stretchout line the spacings in the raked 
profile, 3' to 28', taking each space separately as they 
all differ. Draw parallel lines through these points, 
3" to 28", on the stretchout line F G. These fines nat- 
urally, are parallel with the raked or pitched lines of tfie 
side as shown. 

Project lines from the points in the front profile, 
parallel to the stretchout line F G. These projector 
fines are to intersect like numbered fines as for instance, 
line from point 10 would intersect the line from point 10" 
on line F G giving point 10'", and so on. A line drawn 
through these points of intersection will give the miter 
cut of side moldings to the front molding. 

For the butt miter against the wall, project lines up 
from the points where the rake or pitch fines of the side 
molding hit the wall line; like point 10 ZI on the wall line 
is projected up to line from point 10" on line F G giving 
point 10 I2Z , and so on. A line drawn through these 
points of intersection gives the butt miter cut of the .side 
molding against the wall of the building. The miter 
cuts of the side molding would be extended apart to 
the required length of the molding and laps provided 
on the miter cut for the miter with the front molding. 
There would be two side moldings, bent right and 
left. 



INDEX. 



PAGE 

A 

Abacus (Fig. 90) 13 

Acute angle, Geometrical definition 2 

triangle, Geometrical definition 3 

Alphabetical list of architectural terms 15, 16 

Altitude of a pyramid, Geometrical definition 8 

of a triangle, Geometrical definition 3 

Angle, Comparison between butt miter and miter between 

two moldings at any 76 

Elbow 13 

Finding true at hips of irregular square box 477 

Finding true in compound offsetting elbows in piping. . . . 147 

Geometrical definition 2 

in plan, From profile of horizontal molding to obtain pro- 
file of inclined molding mitering with horizontal molding 

at octagon angle 205 

in plan, From profile of inclined molding to establish pro- 
file of horizontal molding to miter with it at octagon 

angle 266 

To divide into two equal parts 39 

To trisect given 39 

Angular pediment (Fig. 81) 11 

Anvil 162 

Apex, Geometrical definition 3 

Arc compared with chord (Fig. 260) 85 

given, To find center 40 

given, To find center by use of square 90 

of circle, To erect perpendicular to without recourse to 

center 41 

of circle, To strike, by triangular guide, chord and height 

given 41 

To find center, chord and height of segment of circle given. 40 

Arch, curved molding blank, in a circular wall 418 

in circular wall, soffit 416 

in circular wall, splayed 414 

in circular wall, splayed elliptical 412 

semicircular wall, soffit development 408 

Semi-elliptical — Architectural definition 11 

To draw joint lines of elliptical 67 

Architectural terms 10 

Alphabetical list 15, 16 

Architrave, definition 10 

Article, Base rectangular and top elliptical 330 

Base rectangular and top round 328, 331 

Base round and mouth elliptical 381 

Base square, Octagonal 188 

Bottom rectangular and top oblong, semicircular ends 310 

One end rectangular and the other round 337 

Rectangular rounded corners, ends flaring more than sides. 316 

Bound to ellipse offset 452 

Round to oblong, circle wider than narrow side of oblong. 328 

Round to oblong, top and bottom placed centrally 327 

Round to octagon 455 

5 



PAGE 

Article, Round to square, top not central over base 331 

Bound top, elliptical base, one side vertical 345 

Round top, elliptical base, top and base central 343 

Round top, elliptical bottom, top and bottom not parallel.. 364 
Bound top, oblong bottom, semicircular ends, one side 

vertical and one with offset 348 

Round top, quadrant base 354 

Square, inclined base, top round 333 

Top and bottom round, one side vertical 341 

Top and bottom round, small end at one side of center line 339 

Top oblong with semicircular ends, bottom a rectangle. . . . 310 

Top rectangular, base round, base larger than top 335 

Top round, base square and inclined 333 

Top round, bottom oblong, semicircular ending, Two cases. 312 
Top round, bottom oblong, semicircular ends, top and bot- 
tom central 346 

Automatic stokers or spiral conveyor, Revolving 482 

Automobile dust pan 420 

fender, Special design 462 

Axes of given ellipse, To find 66 

Axis, Geometrical definition 9 

B 

Ball, miter between moldings of adjacent gables on square 

shaft formed by a 176 

Fitting against octagon shaft 122 

Fitting against square shaft 122 

To construct in any number of pieces of the shape of gores 124 

Balustrade pedestal, Architectural definition 11 

Bar, Common, single or double pitch skylights, at right angles 138 

Hip, in hipped skylight 219 

Jack, in hipped skylight 221 

Base a quadrant, top round, transitional object 354 

elliptical, top round, top and bottom not parallel 364 

elliptical, top round, transitional article, both central 343 

elliptical, top round, transitional object one side vertical. . 395 

for chimney stack 327 

Geometrical definition 3 

molded, in which projection of sides differs from ends... 182 
oblong, semicircular ends, top round, transitional object, 

straight back, and one with offset 349 

oblong, semicircular ends, round top, transitional object, 

top and bottom central 346 

oblong and top square, flaring article 177 

of column, definition 11 

of finial of curved profile, square in plan, mitering over 

peak of gable coping with a double wash 169 

of octagon finial 123 

of pedestal, definition 1 1 

rectangular and inclined, top round, transitional object. . . 333 

rectangular, top elliptical, transitional object 330 

rectangular, top round, transitional object, top wider than 

narrow side of rectangle 328 



19 



520 



Index. 



PAGE 

Base round, mouth elliptical, Ship ventilator 381 

round, top square, transitional object base larger than top. 335 

square, top octagonal, Tapering article 188 

Bath, Hip 318 

Bath of regular flare, Hip 273 

Bathtub 366 

Flaring, tapering ends, semicircular head, head having more 

flare than sides 314 

Lining head of 378 

Beam compasses or trammels, description 23 

Bed course, Architectural definition 13 

Bed molding, Architectural definition 13 

Bell, elliptical, base round, Ship ventilator 381 

Bifurcated pipe, the arms leaving at same angle, having same 

diameter as main pipe 137 

Blade of spiral conveyor, Revolving 486 

Blank for curved molding 254 

for curved molding in circular wall arch 41S 

Block, Head, Architectural definition 12 

Stop, Architectural definition 12 

Blower connection of irregular shape 503 

for grate ISO 

Offsetting 507 

Boat, Cracker, ornamental fan 267 

Boards, Drawing, Description 18 

Boiler cover with round corners, Raised 309 

with semicircular ends 258 

Locomotive, Blaring section of 179 

Boot, Double offset, round to rectangular 443 

for furnace pipe, bottom oblong, semicircular ends, round 

top, both central 346 

for furnace pipe, bottom oblong, semicircular ends and top 

round, straight back 346 

for furnace pipe, offsetting, bottom oblong, semicircular 

ends, top round 349 

Furnace, offsetting round to oblong with semicircular ends. 393 

Boss, Conical 271 

Faucet, for cans 434 

fitting over molding miter 120 

to fit around faucet 379 

to fit around faucet of can 181 

Bosses for sheet metal hand pumps 448 

Box, Cold air, inclined portion to meet horizontal obliquely. 214 

Cold air, inclined portion joining the level 210 

Irregular shaped 477 

of instruments for drawing, description 28 

Bracket, Architectural definition 12 

Diagonal, under cornice of hipped roof 198 

molding, Architectural definition 13 

ornament 119 

Eaking 193 

Raking, for curved pediment 356 

Raking, raised panel on face 196 

Raking soffit 50S 

To draw scroll to specified width for 69 

Branch at other than right angle of two pipes 151 

at other than right angles of two pipes intersecting at other 

than right angles ' 154 

at right angles between pipes of different diameters 153 

between elliptical and round pipe of larger diameter, at 

other than a right angle, two cases 155 

in tapering pipe, axes crossing at right angles 292 

leaving in two directions. 501 



PAGE 

Branch of pipe tapping into a four-piece elbow 159 

or T-joint at other than right angles between pipes of 

different diameters, axis of smaller at one side of larger 158 
on tapering pipe, axis of branch at other than right angle 
to axis or side of tapering, on taper-joint, axes being 

at right angles 300 

on tapering pipe, axis of branch piece at right angles to 

side of taper joint 296 

or T-joint between pipes of different diameter and at right 
angles, the axis of the smaller pipe passing to one side 

of the larger 156 

or T-joint between two pipes of same diameter • . . 139 

pipe tapping reducing joint off the top 323 

Square pipe, from round main 146 

Tapering, intersected at lower end by main pipe, axis 

crossing at other than right angle 290 

Tapering, intersected at lower end by main pipe, axis 

crossing at right angles 270 

Three-pronged fork with tapering 391 

Unusual two-way Y 429 

Y, arms same diameter as main and leaving at same angle 137 

Y, Double offsetting 466 

Y, of two tapering pipes joining larger pipe at an angle. . 3S9 

Branches, Y, forty-five degrees to main pipe 505 

Breast and spout for cans 302 

Breeches or Y-branch, arms being same diameter as main and 

leaving at same angle 137 

Breeching, Double offset Y-branch 466 

Oblique offsetting Y-branch 310 

Bnusual two-way Y 429 

Y-branch of two tapering pipes joining larger pipe at an 

angle 389 

Broken pediment, definition 12 

To obtain profile and patterns of returns at top and foot 

of segmental 204 

Butt miter against curved surface 99 

against irregular or molded surface 103 

against plain surface, oblique in elevation 97 

ami miter between two moldings at any angle, comparison 76 

Architectural definition 15 

of molding inclined in elevation, oblique in plan 173 

C 

Can boss to fit around faucet 181, 379 

breast and spout 302 

for faucet 271 

Canopy, on range hood, with reduced sides 512 

or marquise for entrances 514 

Cap, curved moldings for elliptical wind 256 

of a pedestal, Architectural definition 11 

Simple curved moldings for window 255 

Square to round base, for chimney 327 

Capital, Architectural definition 13 

Construction of volute for 142 

of a column, Architectural definition 11 

Cavalier projection, Architectural definition 14 

Center, Geometrical definition 4 

of arc, To find when chord and height of segment of circle 

is given 40 

of given arc, To find 40 

of given arc, To find, by use of square 40 

Centers, three sets, To draw approximate ellipse with com- 
passes to given dimensions 65 



Index. 



521 



PAGE 

Centers, To draw spiral from, with compasses 68 

To find in a given ellipse by which approximate figures 

may be constructed 66 

two sets, To draw approximate ellipse with compasses to 

given dimensions 64 

Champfer, Octagon to square, transition piece 191 

Octagon to square mold, transition piece 192 

Chimney top 327 

Chord and height of a segment of circle given, To find 

center from which arc may be struck 40 

and height of circle given, To strike the arc of, by tri- 
angular guide 41 

compared with its arc 85 

Geometrical definition 5 

Chute, Mason 's 464 

Circle, General rule for inscribing regular polygons within . . 44 

Geometrical definition 4 

given, To inscribe regular dodecagon within 44 

given, To inscribe regular nonagon within 44 

or arc of circle, To draw a tangent at a given point 

without recourse to center 41 

To divide into eight equal parts by use of a 45 degree 

triangle : 50 

To divide into eight equal parts by use of 22 i Ax67 1 /j 

degree triangle 51 

To divide into four equal parts by use of triangle or 

set-square 50 

To divide into twelve equal parts by use of 30 x 60 degree 

triangle. 50 

To draw an octagon about 56 

To draw a square about 57 

To draw ellipse as oblique projection of 61 

To draw equilateral triangle about 55 

To draw hexagon about 55 

To draw through any three points not in a straight line. . 41 

To ellipse offset, transition piece 452 

To inscribe a dodecagon within 59 

To inscribe a regular decagon within 44 

To inscribe a regular polygon of eleven sides (undecagon) 

within by general rule 45 

To inscribe a square within 43 

To inscribe an equilateral triangle within 51 

To inscribe equilateral triangle within 43, 57 

To inscribe hexagon within 52 

To inscribe octagon within 53, 58 

To inscribe regular hexagon within 43 

To inscribe regular heptagon within 43 

To inscribe regular undecagon within 44 

To inscribe regular octagon within 43 

To inscribe regular pentagon within 43 

To inscribe square within 52, 5S 

Circles, Two, and intersecting lines of given dimensions, To 

construct ellipse with 63 

and their properties, Geometry of 4 

Circular broken pediment, to obtain modified profiles of re- 
turns at foot and top 204 

cylinder, Geometrical definition 8 

wall, Blank for curved molding in arch in 418 

Soffit of arch in 416 

Soffit of semicircle arch in two cases 408 

Splayed arch in 414 

Splayed elliptical arch in. 412 



PAGE 

Circumference, Geometrical definition 4 

of ciicle or part of circle, To draw straight line equal 

in length to 42 

of given circle, To ascertain 41 

Circumscribed polygons, Geometrical definition 5 

Cloth for drawing, Tracing, description 28 

Coal hod funnel 399 

Cold air box in which inclined portion joins level portion 

obliquely in plan — two solutions 210 

joining furnace at an angle on a central line 147 

to meet horizontal portion obliquely in plan 214 

Collar, Elliptical, tapping wire of larger diameter at other 

than right angle, two cases 155 

intersecting a transformer 488 

of piping intersecting main of larger diameter at right 

angles 153 

intersecting main pipe of same diameter at right 

angles 139 

intersecting main pipe at other than right angle 154 

tapping main pipe of larger diameter at other than 
right angle axis of collar being to one side of main 

pipe 158 

tapping main pipe of larger diameter, at right angles, 

axis of collar passes to one side of main pipe 156 

of round piping tapping four-pieced elbow 159 

on taper joint, axes crossing at right angles 292 

axis of collar at other than right angle to axis or 

side of taper-joint 298 

axis of collar being at right angles to side of taper 

joint 296 

Flaring, round top and bottom, placed obliquely to each 

other 361 

round top, square bottom, to fit around pipe passing 

through inclined roof 333 

Round, tapping main pipe at angle in two directions 501 

Tapering, intersected at lower end by main pipe, axes 

crossing at right angles 270 

intersecting the main pipe at other than right angle .... 290 

on taper-joint, intersecting obliquely 302 

on taper-joint, axes being at right angles 300 

straight back, reducing joint at top 323 

to join two pipes of unequal diameters at an angle 359 

Collars, Two round, tapping main pipe at forty-five degrees. 505 

Column, Architectural definition 11 

Common bar for skylights 138 

Compasses and dividers, description 22 

use of for drawing approximate ellipses 64 

use of to draw spiral from centers 68 

Complement, Geometrical definition 6 

Compound curve rectangular shaped elbows 494 

curved elbow for square 478 

in square pipe 140 

offset in rectangular pipe 210, 214 

offsetting elbows in round piping 147 

Concave, Geometrical definition 10 

Concentric circles, Geometrical definition 5 

Concrete, Mason's chute for 464 

Conductor or roof leader pipe, compound curved elbow for 

square 478, 494 

Cone and frustum, Envelope of, described by pin and thread 80 

base an elliptical figure, Envelope of the frustum of 260 

base a true ellipse, Frustum of 317 



522 



Index. 



PAGE 

Cone, base an approximate ellipse, upper plane of frustum 
oblique to axis, tapering article with equal flare through- 
out corresponding to frustum of 276 

base oblique to its axis, E'nvelope of right 265 

cut by a plane parallel to its axis, Envelope of a right. . . . 265 
contained between planes oblique to its axis, Envelope of 

frustum of right 274 

Envelope of an elliptical 308 

of a right 249 

of frustum of right 250 

of scalene 306 

of frustum of scalene 311 

fitting against a surface of two inclinations, Frustum of. . 269 
Frustum of, intersecting a cylinder of greater diameter 

than itself at other than right angles 290 

deriving the pattern from the drawing 82 

deriving the pattern of from wooden model 82 

Geometrical definition 9 

right 80 

triangulated, elliptical 92 

Geometrical definition 8 

having irregular base, Envelope of frustum of elliptical. . 322 

hopper intersected by spout 492 

intersected at its lower end by a cylinder, axes intersecting 

at right angles, Frustrum of 270 

intersected by cylinder at its upper end, axes crossing at 

right angles 275 

intersected by a cylinder of less diameter than itself at 

right angles to its base 294 

intersected by cyliuder of less diameter than itself, axes 

crossing at right angles 292 

intersected obliquely by a cylinder, axes not lying in same 

plane, Frustum of a scalene 323 

in which axis of cylinder not at right angles to axis nor to 

side of cone, Cylinder joining frustum of 298 

joining a cylinder of greater diameter than itself, axis of 
frustum passing to one side of cylinder, Frustum of.... 291 

Method of obtaining the lines upon elevation 81 

of greater diameter than itself at right angles, Cyliuder 

joining 296 

perspective view for tiangulating illustration, Frustum of 

elliptical 92 

plan and elevation for constructing a wooden one for pur- 
poses of illustration 81 

Badial line triangulating frustum of elliptical 92 

Eight, generated by the revolution of a right-angled tri- 
angle about its perpendicular 79 

Eight, with thread fastened at apex to which are attached 
points for marking upper and lower base of frustum. ... 79 

Eight, Unfolding envelope of 80 

Scalene, System of triangulation for 94 

upper plane oblique to its axis, Envelope of frustum of right 265 

To draw ellipse as a section 62 

Triangulated elliptical 91 

truncated obliquely 81 

Cones of unequal diameter intersecting at right angles of 

their axes 300 

joined and pin with string attached at apex of larger 
sector, third step in constructing solid of elliptical flar- 
ing article 85 

of unequal diameters intersecting obliquely, Frustums of. 302 
parts joined and completed, final step in constructing solid 
of an elliptical flaring .article 85 



PAGE 

Cones, Sectors of small, second step in constructing solid of 

elliptical flaring article 84 

Conical boss 271 

development or flaring work 79 

hip batch, Eegular flare 273 

lip for sheet metal pitcher 272 

method for rectangular oblique offsets 494 

method for rectangular oblique twisted offset 497 

roof flange to fit around pipe and against roof of one 

inclination 266 

to fit around pipe and against roof of double inclina- 
tion 269 

scale scoop, both ends alike 278 

spire mitering upon eigh* gables 282 

mitering upon four gables 280 

system for octagonal semicircular pipe 253 

two-piece elbow 284 

Conic section, Geometrical definition . . . . • 9 

Constructing ellipse to given dimension by two circular and 

intersecting lines 63 

equilateral triangle upon given side 45 

regular polygon of thirteen sides by general rule, length 

of side being given 48 

triangle, length of sides being given 45 

Construction of regular polygons by protractor 57 

by compasses and straight edge 43 

by T-square and triangles or set-squares 49 

Convex, Geometrical definition 10 

Conveyor or loaded spout, Spiral 426 

Eevolving blade for spiral 482 

Corbel, Architectural definition 12 

Corner piece of mansard roof 105 

Corners, rounded, for regular flaring oblong article 258 

Cornice, Architectural definition 10 

Deck, Architectural definition 12 

Lintel, Architectural definition 12 

Co-secant, Geometrical definition 7 

Co-sine, Geometrical definition 6 

Co-tangent, Geometrical definition 6 

Cover with round corners, Eaised boiler 309 

with semicircular ends, Oblong raised 258 

Cowl or ship ventilator, base round, mouth elliptical 381 

Cracker boat 267 

Cripple or jack bar of hipped skylight 221 

Crown molding, Architectural definition 13 

Cube, Geometrical definition 8 

Curved elbow, Compound for square pipe 478 

line, Geometrical definition 1 

mansard roof, Hip finish in 101 

square base, octagonal top, Fascias of a hip molding 

finishing against 230 

molding, Blank for 254 

in an elliptical window cap 256 

in arch in circular wall, Blank for 418 

in window cap 255 

meeting straight molding of same profile, Obtaining 

miter line of 152 

pediment, Baking bracket for 356 

surface, Double, Geometrical definition 3 

Single, Geometrical, definition 3 

Curves for molding covering hip of curved mansard roof, to 

obtain 238 

Cylinder cut off obliquely 125 



Index. 



523 



PAGE 

Cylinder of different diameters intersecting at other than right 

angles 154 

Frustum of scalene cone intersected obliquely by 323 

Geometrical definition 8 

intersected by cone of smaller diameter at its lower end, 

at other than right angles 290 

intersected by an elliptical one of smaller diameter at 

other than right angles. Two cases 155 

intersected by another of smaller diameter at right angles, 

axis of smaller passing to one side 156 

intersected by another of smaller diameter at other than a 

right angle, axis of smaller cylinder being to one side of. 158 
intersected by frustum of cone of smaller diameter at right 

angles, axis of frustum passing to one side 291 

intersected obliquely by a rectangular pipe 146 

intersecting cone of greater diameter, axes crossing at 

right angles 292 

intersecting double plane 127 

intersecting another of larger diameter at oblique angles. . 501 

intersecting another of larger diameter at right angles. . 153 
intersecting another of same diameter at other than right 

angle 151 

intersecting another of same diameter at right angles. . . . 139 

intersecting irregular shaped object 488 

intersecting miter of a molding 120 

intersecting lower end of frustum of cone, axes intersecting 

at right angles 270, 271 

joining a cone of greater diameter than itself 296 

joining frustum of cone, axis of cylinder not at right angles 

to axis nor to side of cone 298 

mitering with peak of gable having double wash 171 

penetrating a four-piece elbow 159 

To draw ellipse shape of oblique section of 61 

True angles in compound intersecting 147 

Cylinders, two, intersecting larger cylinder at forty-five de- 
grees 505 

D 

Decagon, Geometrical definition 4 

patterns for newel post, plan of which is a 117 

regular, to draw upon a given side 47 

regular, to inscribe within a given circle 44 

Deck cornice, Architectural definition 12 

molding, Architectural definition 12 

Definitions and terms, Glossary of geometrical 1 

Degree, Geometrical definition 5 

Dentil, Architectural definition 12 

band, Architectural term 10 

course, Architectural term 10 

molding, Architectural term 10 

Descriptive geometry, Relationship to pattern cutting 1 

Detail drawing. Architectural definition 14 

Development of surfaces, Geometrical definition 15 

Diagonal bracket under cornice of hipped roof 198 

Geometrical definition 4 

scale rule for drawing, description 25 

Diagram of sections for triangulation 92 

Diameter, Geometrical definition 5 

Diamond faucet boss for cans 434 

shaped frustum of pyramid 178 

Die of pedestal, Architectural definition 11 

Dividers and compasses, description 22 

Dodecagon, drawing upon a given side 48 



PAGE 

Dodecagon, Geometrical definition 4 

regular, to inscribe within a given circle 44, 59 

urn, plan 118 

Dodecahedron, Geometrical definition 9 

Dormer, Molding on side of _. . . . 165 

Double curved surface, Geometrical definition 3 

elbow joining two other pieces not in same plane, Inter- 
mediate piece of 147 

offset in rectangular pipe 210, 214 

Y-Branch 466 

Drafting, Pattern, Exemplification of principles of 71, 95 

tal iles, description 17 

terms and definitions 14 

Drawing approximate ellipse with compasses, length only 

given 64 

with compasses to given dimensions, using three sets 

of centers 65 

with compasses to given dimensions, using two sets 

of centers 64 

boards, description 18 

circle through any three given points not in straight line. 41 

detail or working, Architectural definition 14 

ellipse to given dimensions by square and strip of wood. . 61 

to given dimensions by a trammel 60 

within given rectangle by means of intersecting lines. . 63 

elliptical flaring article by the usual method 83 

equilateral triangle about given circle 56 

upon a given side 53 

hexagon about a given circle 55 

ionic volute 67 

joint lines of elliptical arch 67 

Linear 30, 35 

line parallel to another by triangles or set-squares 36, 37 

Method of deriving pattern of frustum of cone from 82 

Obtaining envelope of molding from, by T-square 75 

octagon about a given circle 56 

pins or thumb tacks, description 27 

regular decagon upon a given side 47, 48 

regular hexagon upon a given side 46, 54 

regular heptagon upon a given side 46 

regular nonegon upon a given side 47 

regular octagon upon a given side 47, 54 

within a given square. . . . , 49 

regular pentagon upon a given side 46 

regular polygon, length of side given, by a general rule. . 48 

regular undecagon upon a given side 4S 

Scale, Architectural definition 14 

scroll to specified width for bracket or modillion 69 

simple volute 67 

spiral for centers with compasses 68 

with spool and thread 69 

square about a given circle 57 

square upon a given side , 53, 57 

straight line equal in length to the circumference of a circle 

or part of circle 42 

parallel to a given line 36 

tangent to circle, or arc of circle at a given point without 

recourse to center 41 

tracing paper and tracing cloth, description 28, 29 

tools and materials, Discussion of 17 

Drip, Architectural definition 13 

Drop upon the face of a bracket 119 

Dust pan for automobiles 420 



524 



Index. 



PAGE 

Dust pipe, Compound curved elbow for 47S 

Conical method for oblique offset 494 

Conical method for oblique twisted offset 497 

Offsetting blower connection 507 

separator, Spiral strip for, short rule 496 

Spiral strips in 432 

work, Compound elbow in square pipe 140 

E 

Eccentric circles 5 

Edges, Straight, description 19 

Egg-shaped flaring pan 263 

Eight-sided tinial base 123 

pedestal 115 

Elbow, Compound curve 478 

Conical method for octagonal shaped pipe 253 

Five-piece 133 

for five pieces, Regular tapering 287 

Four-piece 132 

intersected by round pipe through a miter 159 

tapering 440 

in elliptical pipe, Two-piece 130 

in square pipe, Compound curved 140, 494 

Oblique twisted offset 497 

Intermediate piece of double, joining two pieces not in 

same plane 147 

in tapering pipe, Three-piece 2S5 

Two-piece 284 

Oblique offset, in rectangular pipe 210, 214 

Offsetting, round to oblong, semicircular ends 482 

one end round, the other elliptical, Eight angle two-piece. 387 

problem in furnace fittings, Reversible 472 

Eight angle piece, to connect round with rectangular pipe. 405 

Eound to elliptical three-piece 453 

Eound to rectangular 437 

Bound to rectangular, transition piece 431 

Square to round seven -piece 405 

Square, in elliptical pipe, two cases 130 

Square, in round pipe, Two-, three-, four- and five-piece. 131, 133 

Square transforming 473 

Three-piece, to join round with elliptical pipe 403 

with middle piece gored 187 

Two-piece 130 

Two-piece, circle to ellipse offsetting 452 

Elbows, square or rectangular pipe 469 

of two smaller pipes, Junction of large pipe with 384 

Elevation, Drawing, Architectural definition 14 

Eleven-side regular polygon (undeeagon), To inscribe within 

a given circle 45 

Ellipse, Approximate, Drawing with compasses to given di- 
mensions, using three sets of centers 65 

Drawing with compasses to given dimensions, using two 

sets of centers 64 

To draw with compasses, length only given 64 

Finding true axes 66 

Geometrical definition 7 

treatise 59, 67 

given, To find centers so that an approximate figure may 

be constructed 66 

of given dimensions, To draw with square and strip of wood 61 

of given dimensions, To draw with a trammel 60 

of specified dimensions, To draw with a string and pencil. 60 



PAGE 

Ellipse, To construct, to given dimensions by two circles and 

intersecting lines 63 

To describe shape of an oblique section of cyilinder, or as 

oblique projection of circle 61 

To draw, as section of a cone 62 

To draw, within a given rectangle by intersecting lines. . 63 

Elliptical arch, To draw joint lines of 67 

base and round top, base and top central 343 

one side vertical 345 

top and bottom not parallel 356 

cone, Envelope of 308 

having an irregular base, Envelope of a frustum of.... 322 

system of triangulation 91, 93 

cylinder, Geometrical definition 8 

flaring article 260 

Circling cones and their sectors to obtain solid of. . 84, 85 

corresponding to frustum of cone 317 

Describing envelope with pin and thread 85 

top placed eccentrically to bottom 320 

Usual method of drawing 83 

frustum of cone, upper plane of frustum oblique to axis. . 276 

mouth and round base, Ship ventilator 281 

pipe and round pipe of larger diameter at other than right 

angles, Joint between two cases 155 

flange to fit against, when pipe passes through center of 

pyramid or hipped roof 145 

to fit against roof 125 

splayed arch in circular wall 412 

to circle offset transition piece 452 

to round pipe offset 396 

three-piece elbow 403 

two-piece elbow Z 387 

rectangular base, transitional object 330 

two-piece elbow, two cases 130 

vase constructed in twelve pieces 183 

window cap, Curved molding for 256 

End, Flaring of oblong tub 178 

Engaged column, Architectural definition 11 

Engine hood, French style, for autos 458 

F"ntablature, Architectural definition 10 

Entrance canopies or marquise 571 

Envelope of cone and frustum described by pins and thread. 80 

molding, Obtaining from a drawing by T-square 75 

Obtaining by using lines from model 74 

ordinary elliptical flaring article described by pin and 

thread 85 

right cone, Unfolding 90 

solid, Geometrical definition 9 

Equilateral triangle 3 

pedestal Ill 

To construct upon given side 45 

To draw about given circle 55 

To draw upon given side 53 

To inscribe within given circle 41, 51, 57 

Eraser or rubber for drawing, description 29 

Erecting perpendicular at given point in straight line by 

compasses and straight edge 36 

to arc of circle without recourse to center 41 

E volute 7 

F 

Face and side of plain tapering keystone 104 

Face miter at an angle 108 



Index. 



525 



PAGE 

Face miter at right angles in molding around panel 106 

Square, produced where square return was intended. ... 78 
miters of molding bounding irregular shaped four-sided 

figure . 109 

Fascia, Architectural definition 13 

Faseias of hip molding finishing curved mansard roof, square 

base, octagonal top 236 

Faucet boss for cans 379 

problem 434 

Can boss to fit around 181 

Conical can boss for 271 

Fender, Special design automobile 462 

Figure, Plane, Geometrical definition 3 

Quadrilateral, Geometrical definition 3 

Rectilinear, Geometrical definition 3 

Figures, straight sided, Geometrical terms and definitions. ... 3 

Fillet, Architectural definition 13 

Finial, Architectural definition 13 

base of curved profile, square in plan, centering over peak 

of gable coping having double wash 169 

Irregular polygon 185 

Octagon, with alternate long and short sides 189 

Fitting, Unusual design 455 

Five-piece elbow 133 

tapering elbow 2S7 

Five-prong fork 450 

Five-sided vase 112 

Flag pole, Ornamental roof flange for 475 

Flange, Architectural definition 14 

Conical, to fit around pipe against roof of double inclination 269 

to fit around pipe against roof of one inclination 266 

Flaring round bottom, top to fit pipe passing through in- 
clined roof 363 

for a pyramidal flange to fit against sides of round pipe 

passing through its apex 144 

Ornamental roof, for pipe or flag pole 475 

to fit around pipe passing through inclined roof 369 

to fit against sides of elliptical pipe passing through center 

of pyramid hipped roof 145 

to fit around pipe and over ridge of roof 128 

Flaring article, base oblong, top square 177 

corresponding to frustum of cone whose base is a true 

ellipse 317 

elliptical describing the envelope with a pin and thread. . 85 

Irregular, elliptical at base, round at top 345, 364 

top and bottom round and parallel, but placed eccen- 
trically in plan 311 

with circular top and quadrant base 354 

or transition piece, round top, oblong bottom 346 

Rectangular 104 

with rounded corners 316 

Regular, oblong with semicircular ends 258 

round base, square top 335 

round top and bottom 339 

round top and bottom, one side vertical 341 

top round, bottom oblong, semicircular end, two cases. . . 312 
collar, lower end intersecting main straight pipe at other 

than right angle 290 

lower end intersecting main straight pipe, axes crossing 

at right angles 270 

top and bottom round, placed obliquely to each other, 

pattern for 361 

elliptical article, Combined portions of cones to make 83 



PAGE 

Flaring elliptical article, Obtaining solid of and envelope of 84, 85 

Usual method of drawing 83 

end of an oblong tube 178 

flange, round bottom, top to fit around pipe passing through 

roof 368 

Regular, oblong article, round corners 259 

pan, Oval or egg-shaped 263 

section of locomotive boiler 179 

shape, forming transition from round horizontal base to 

round top, placed vertically 376 

tray, Heart-shaped 261 

tub, tapering sides and semicircular head 314 

two-piece elbow 2S4 

work or regular tapering forms 79 

Flat arch, Architectural term 11 

scale rule for drawing, description 25 

Float, Soap maker 's 320 

Foot molding, Architectural definition 13 

Forge, Hood of portable 371 

Fork, making three-pronged in two pieces 425 

Three-pronged, with tapering brandies 391 

two-prong, consisting of two tapering pipes joining larger 

pipe at an angle 389 

two-pronged offsetting 466 

Unusual five-pronged 450 

Forms, Irregular or triangulation 86, 95 

Parallel, First system of pattern cutting 72 

Four-piece elbow 132 

through a miter, Pipe intersecting 159 

tapering elbow 440 

Four-sided figure, Molding mitering around an irregular 109 

pedestal 112 

Frieze, Architectural definition 10 

French style auto hood 458, 461 

Frustum and cone described by pin and thread, Envelope of. . SO 

Envelope of right cone 250 

of cone, base an approximate ellipse, upper plane of frus- 
tum oblique to axis, Tapering article with equal flare 

corresponding to 276 

base an elliptical figure, Envelope of 260 

deriving pattern 'from drawing 82 

deriving pattern from wooden model 82 

fitting against surface of two inclinations 269 

Geometrical definition 9 

intersected at its lower end by cylinder 270 

intersected by cylinder, axis neither at right angles nor 

side of frustum 298 

intersecting cone of larger diameter at right angles 

to axis 300 

intersecting cylinder of greater diameter, at other 

than right angles 290 

joining cylinder of greater diameter at other than right 

angles 291 

Using right cone with a thread fastened at apex, 

points marking upper and lower base of 79 

of elliptical cone having irregular base, Envelope of 322 

octagonal pyramid, Envelope of 243 

having alternate long and short sides 244 

pyramid, Envelope of, diamond shaped 178 

Geometrical definition 9 

of right cone contained between planes oblique to its axis, 

Envelope of 274 



526 



Index. 



PAGE 

Frustum of right cone, lines showing portion of cone removed 

to produce it 80 

upper plane oblique to its axis, Envelope of 265 

of scalene cone, Envelope of 311 

intersected obliquely by a cylinder 323 

of square pyramid, Envelope of 242 

Frustums of two cones of unequal diameters intersecting 

obliquely 302 

Funnel end scale scoop 279 

coal hod 399 

furnace bonnet, Collar intersecting axes crossing at right 

angles 292 

Collar intersecting, axis of collar at right angles to side 

of bonnet 296 

Collar intersecting axis of collar neither at right angles 

to axes nor to side of bonnet 298 

boot, Double offsetting .' 443 

elbows, Reversible fitting 472 

fitting for quarter turn in pipe, oblong, semicircular ends. . 350 
oblong, semicircular ends at bottom, round top, back 

straight, two cases 348 

offsetting, round to oblong, semicircular ends 393 

rectangular bottom and oblong, semicircular ends at top. 310 

round to oblong, with semicircular ends, two cases 312 

pipe, oblong, semicircular ends, Taper-joint for 258 

G 

Gable, Architectural definition 12 

copmg, Cylinder mitering with peak of, with double wash. 171 
square shaft of curved profile, mitering over peak of 

gable coping double wash 169 

cornice mitering upon inclined roof 164 

miters, Simple 110 

molding, Architectural definition 15 

From profile of inclined molding to establish profile of 

horizontal molding to miter with it at octagon angle.. 206 

mitering against a molded pilaster 161 

at octagon angle to horizontal molding, obtaining modi- , 
fied profile in inclined molding from normal profile 

in horizontal molding 205 

or broken pediment, To obtain profile of horizontal return 

at top of 202 

or pediment with modified profile and normal profile in 

horizontal return 201 

or pediment with normal profile and modified profile in 

horizontal return 200 

moldings and roof pieces of octagonal pinnacle 174 

and roof pieces of square pinnacle 173 

Gables, Adjacent, having different pitches upon an octagon 

pinnacle, Miter between moldings of 20S 

Miter between moldings of, with different pitches, upon 

pinnacle with rectangular shaft 207 

Eight, Conical spire mitering on 282 

Octagon spire mitering upon 245 

Four, Conical spire mitering on 280 

Octagon spire mitering on 246 

Square spire mitering on 244 

General rule for inscribing any regular polygon within given 

circle 44 

Geometrical problems 36, 70 

Geometry, definition 1 

Globe or sphere, Geometrical definition 9 

Glossary, Architectural and geometrical 15 



PAGE 

Goose-neck square roof leaders, Compound curved elbow 478 

with compound curve 494 

Gore piece forming transition from octagon to a square. . . . 191 
Molded article forming transition from a square to an 

octagon 192 

three-piece elbow, middle piece being a gore 187 

Gores, To construct ball in any number of pieces of the 

shape of 124 

Grate blower ISO 

Grocers' display receptacle for coffee 273 

scale scoop, funnel shaped end 279 

Guard, Automobile, Special design fender 462 

Gusset piece in three-piece elbow 187 

H 

Head block or truss, Architectural definition 12 

for curved or inclined molding butting against oblique flat 

surface, To find true profile of 104 

of bath-tub lining 378 

Heart-shaped flaring tray 261 

Height and chord of circle given, to strike an arc of circle 

by triangular guide 41 

Heptagon vase 115 

Heptagon, regular, To draw upon a given side 47 

To inscribe within a given circle 43 

Hexagon, Geometrical definition 4 

pedestal 114 

regular, To draw on given side 46, 54 

To inscribe within a given circle 43 

To draw about given circle 55 

To inscribe within given circle 52 

Hexagonal prism, Geometrical definition 8 

pyramid, Envelope of 241 

Hexahedron, Geometrical definition 9 

Hip, Architectural definition 14 

bar, top and bottom of 219 

bath 318 

Regular flare 273 

finish, Curved mansard roof 101 

mold on octagon angle in mansard roof mitering against 

bed molding of corresponding profile 223 

molding, Architectural definition 13 

bottom of on mansard roof, octagon at top, square at 

bottom 233 

covering curved mansard roof 238 

mitering against bed molding of deck cornice on man- 
sard roof square at base, octagonal at top 230 

mitering against planceer of deck cornice on mansard 

roof, square at eaves, octagon at top 227 

octagon angle of mansard roof mitering upon inclined 

wash at bottom 225 

on right angle in mansard roof mitering against planceer 

cf deck cornice. .'. 215 

Eight angle, in mansard roof mitering on bed molding. . 217 
Straight, corner piece of mansard roof, embodying prin- 
ciples upon which mansard finishes are developed.... 105 
Hipped roof, Flange to fit against sides of elliptical pipe 

passing through center of 145 

Pipe passing centrally through apex of 144 

Hod, Funnel coal 399 

Hole in roof for pipe to pass through 105 

Hood, French style auto 458 

Oil lank 374 



Index. 



527 



PAGE 

Hood, Portable forge 371 

Range, with reduced sides 512 

Hopper 104 

Cone shaped, intersected by spout 492 

or flaring article, base oblong, top square 177 

or irregular shaped box 477 

Horizontal lines, Geometrical definition 2 

molding, to miter with inclined molding at an octagon 
angle, and patterns for both arms, To establish profile 

of 206 

To obtain profile of inclined molding to miter with it at 

octagon angle and patterns for both arms 205 

return at foot of gable to miter at right angles with 

inclined molding of normal profile 200 

at top of broken pediment to miter with inclined mold- 
ing, To obtain the profile of 202 

Horseshoe arch, Architectural term 11 

House, Architectural diagrams of 14 

Hyperbola, Geometrical definition 7 

Hypothenuse, Geometrical definition 3 

I 

Icosahedron, Geometrical definition 9 

Import, Architectural term 11 

Inclined molding, From the profile of a given horizontal 
molding to obtain the profile of inclined molding neces- 
sary to miter with it 205 

mitering upon wash, including return 166 

to miter at right angles with horizontal return, To obtain 

profile of 201 

or oblique lines, Geometrical definition 2 

roof, Gable cornice mitering upon 164 

Incised work, Architectural definition 13 

India rubber for drawing, description 28 

Ink, India, for drawing, description 27 

slab for drawing, description 27 

Inscribed circles. Geometrical definition 5 

polygons, Geometrical definition 5 

Inscribing dodecagon within given circle, 59 

eleven-sided regular polygon (undecagon) within given 

circle by general rule 44 

equilateral triangle within given circle 43, 51, 57 

hexagon within given circle 52 

octagon within given circle 53 

regular decagon within given circle 44 

dodecagon within given circle 44 

heptagon within given circle 43 

hexagon within given rule 43 

octagon within given circle 43 

within given square 49 

pentagon within given circle 43 

polygon within given circle by general rule 44 

undecagon within given circle 44 

square within given circle 43, 52, 58 

Inside miter, Architectural definition 15 

Instruments for drawing, description 28 

Intersecting elbows, Junction of large pipe with elbows of 

smaller pipes 384 

lines and circles, To construct ellipse of given dimensions. . 63 

to draw ellipse within given rectangle by 63 

Intersection of solids, Geometrical definition 9 

Involute, Geometrical definition 7 

Ionic volute, To draw 67 



PAGE 

Irregular forms of triangulation 86, 95 

four-sided figure, Face miter patterns for molding bounding 109 

or mold surface, Butt miter against 103 

shaped article, illustrating principles of triangulation 87 

Isometrical projection, Architectural definition 14 

Isosceles triangle, Geometrical definition 3 

J 

Jack bar pattern for top of skylight 221 

Joint at other than right angles between two pipes of dif- 
ferent diameters 151 

between an elliptical and round pipe of larger diameter. . 125 
two pipes of different diameters intersecting at other 

than right angles 551 

two pipes of same diameter at other than right angles.. 415 

lines of elliptical arch, To draw 67 

T -joint between pipes of same diameter 139 

Junction of large pipe with elbows of smaller pipes of same 

diameter 3S4 

K 

Keystone, Architectural term 11 

having molded face with sink 120 

L 
Level molding mitering obliquely against level molding of 

different profile, pattern for 167 

Lead pencils for drawing, description 25 

Leaders, Square, Compound curved elbows for, conical 

method 494 

Compound curved elbows for 478 

Line, Geometrical definition 1 

parallel to another. To draw by triangles or set-square.... 36 

perpendicular to another, To draw by triangle or set squares 37 

Straight, To divide into equal parts by scale 38 

To divide into two equal parts by a pair of dividers. ... 37 

To divide into two equal parts with compass 37 

To divide into two equal parts by triangle or set square. 38 
To draw equal in length to circumference of circle or 

part of circle 42, 43 

Linear drawing, Principles of 30, 35 

Lines, Joint, in elliptical arch, to find 67 

Rise of, in obtaining envelope of molding from model 74 

Straight and parallel to given line and at given distance 

from it, To draw with compass and straight edge 36 

Lining of head of bathtub 328 

Lintel cornice, Architectural definition 12. 

molding mitering obliquely against arch molding of differ- 
ent profile 167 

Lip, Sheet-metal pitcher 272 

Loading, Spiral spout or conveyor 476 

Locomotive boiler, Flaring section of 179 

Lozenge, Rhombus, Geometrical definition 4 

M 
Mansard roof, Corner piece of, embodying principles upon 

which all mansard finishes are developed 105 

Hip finish in curved mansard roof, hip being a right 

angle 101 

Hip molding mitering against bed molding of deck cor- 
nice on mansard roof, square base, octagonal top. . . . 208 
Hip molding mitering against planceer of deck cornice 
on mansard roof, square eaves, octagon top 227 



528 



Index. 



PAGE 

Mansard roof, mitering against bed molding at top, Hip 

molding upon a right angle in 217 

mitering against bed molding of corresponding profile, 

Hip mold upon octagon angle in 223 

mitering against planceer of deck cornice, Hip molding 

upon right angle in 215 

mitering upon inclined wash at bottom, Hip molding 

upon octagon angle of 225 

octagon top, square bottom, Miter at bottom of hip mold- 
ing on 233 

square base, octagonal top, Fascias of hip molding finish- 
ing curved 236 

To obtain curves for molding covering hips of curved.. 238 

Marquise, Construction and patterns 514 

Mason 's chute 464 

Materials and tools for drawing, discussion 17 

Middle piece gored in three-piece elbow 187 

Miter, Architectural definition 15 

between moldings of adjacent gables upon square shaft 

formed by means of a ball 176 

between moldings of different profiles 102 

butt against curved surface 91 

against irregular or molded surface 103 

against plain surface oblique in elevation 97 

against plain surface oblique in plan 97 

cutting or parallel forms, first system of 72 

Requirements for 76 

. Rule for 77 

face at right angles, as in molding around panel 106 

line, Curved molding of same profile, for straight 110 

or jack bar of hipped skylight 221 

return at other than right angle as in cornice at corner of 

building. . .'. 98 

Square, Comparison between the short rule for cutting 

square miter and the method prescribed by rule 77 

Square faced, produced where a square return miter was 

intended 78 

Square return, or miter at right angles 98 

Usual method of cutting 76 

Miters, Simple gable 110 

Model, Cardboard, perspective view illustrating application of 

principles of triangulation 89 

illustrating principles of triangulation 89 

Use of for obtaining envelope of molding by lines 74 

Wooden, Method of deriving pattern of a frustum from. . S2 
Modified profile in horizontal return of broken pediment .... 202 
in horizontal return of raking molding or pediment with 

normal profile 200 

in raking molding or pediment and normal profile in 

horizontal return 201 

profiles for return at top and bottom of segmental broken 

pediment 204 

Modillion, Architectural definition 12 

course, Architectural term 10 

head, Architectural term 10 

land, Architectural term 10 

molding, Architectural term 10 

To draw scroll to specified width for 69 

Molded base, projection of sides different from ends 1S2 

face, Keystone with 120 

or irregular surface, butt miter against 103 

Pilaster, Gable molding mitering against 161 

Molding, Architectural definition 13 



PAGE 

Molding, against octagonal side of lower roof of dormer 

mitering 165 

Crown, etc., Architectural term 10 

Curved blank 254 ' 

Face, miter at right angles 106 

Gable, mitering against molded pilaster 161 

Generating, in plastic material by reverse stay. 73 

Hip, at right angles in mansard roof miter against planceer 

of deck cornice 215 

covering curved mansard roof 238 

finishing curved mansard roof, square base, octagonal top 236 
mitering against planceer of deck cornice on mansard 

roof, square eaves, octagonal ridge 227 

mitering against bed molding of deck cornice on mansard 

roof, square base, octagonal top 230 

miter at bottom of, on mansard roof, octagon top, square 

bottom 233 

upon octagon angle of mansard roof, mitering against 

bed molding of corresponding profile 223 

upon octagon angle of mansard roof mitering upon 

inclined wash 225 

upon right angle in mansard roof, mitering against 

bed molding at top 217 

inclined in elevation against plain surface, Butt miter of. . 173 

in elliptical window cap, Curved^ 256 

mitering around irregular four-sided figure 109 

Obtaining envelope of from model by use of lines 74 

Raked or modified, Architectural definition 14 

Straight, meeting curved molding of same profile, To obtain 

miter line and pattern 152 

Moldings and roof pieces in gables of octagon pinnacle .... 174 

in gables of square pinnacle 173 

bounding panel triangular in shape 108 

Face miter, bounding irregular four-sided figure 109 

for window cap, Simple curved 255 

of adjacent gables on square shaft, miter between formed 

by ball 176 

of different profiles, Miter between 102 

of different profiles mitering at angle 167 

at any angle, Comparison between butt miter and miter. . 76 

Monument, octagon, with alternate long and short sides 189 

Moresque arch, Architectural term 11 

N 

Neck mold, Architectural term 13 

Newel post, Decagon 178 

Ninety-degree elbows in elliptical pipe, two cases 130 

in pipe, two, three, four and five pieces 133 

Nonagon, regular, To draw upon given side 47 

To inscribe within given circle 44 

Norman profile in horizontal return and modified profile in 

raking molding or pediment 251 

in raking molding and modified profile in return 200 

or normal stay, Architectural definition. . 14 

O 

Oblique cone, Geometrical definition 8 

cylinder, Geometrical definition 8 

offset in rectangular piping 210, 214 

offsetting blower connection 507 

Y-Branch 510 

or inclined lines, Geometrical definition 2 

or scalene cones 306 



Index. 



529 



PAGE 

Oblique projection of circle, To draw for an ellipse 61 

section of cone, Describing for an ellipse 62 

of cylinder, To describe form of for ellipse 61 

Oblong base and square top flaring article 177 

pipe with semicircular ends, To join round pipe and offset. 373 

raised boiler cover, semicircular ends 258 

to round double offsetting boot 443 

tub, Flaring end of 178 

vessel, semicircular top, rectangular bottom, End" of 310 

Obtuse angle, Geometrical definition 2 

Obtuse-angled triangle, Geometrical definition 3 

Octagon, Geometrical definition 4 

miter, Architectural definition 15 

pinnacle, miter between moldings of adjacent gables upon. 208 

pipe, Semicircular elbow for 253 

regular, To draw upon given side 47, 54 

To draw within given square 49 

to inscribe within given circle 43 

shaft fitting over ridge of room, Curved 123 

fitting over ridge of roof 126 

mitering upon ridge and hips of roof 127 

to fit against a ball 122 

spire, mitering upon eight gables 245 

mitering upon four gables 246 

mitering upon roof at junction of ridge and hips 248 

square transition piece 455 

To draw about given circle 56 

To inscribe within given circle 53, 58 

to square, gore piece forming transition at end of chamfer 191 

gore piece forming transition in molded article 192 

with alternate long and short sides, patterns of 1S9 

Octagonal pedestal 115 

pinnacle, Moldings and roof pieces in gables of 174 

Pyramid, alternate long and short sides, Envelope of frus- 
tum of 244 

Envelope of frustum of 243 

side of tower, Dormer molding mitering against 165 

top and square base tapering article 188 

Octahedron, Geometrical definition 9 

Offset between two pipes, oblong in section whose long diam- 
eters meet at right angles 350 

blower 507 

boot, Developing double 443 

circle to ellipse transition piece 452 

in ventilation pipe 497 

to jom an oblong pipe with a round one 393 

to join round pipe to one of elliptical profile 396 

twisted, for square pipe 478 

Y, from circular pipe 510 

Offsetting branch 466 

compound elbows in round piping, True angles of 143 

elbows at any angle 136 

elbows for square pipe, oblique 494 

elbows for square pipe, obliquely twisted 497 

elbow, oblong, semicircular ends to round 482 

furnace bottom, round to oblong semicircular ends 349 

square to octagon transition piece 455 

Oil tank-hood 374 

Opening in roof for pipe to pass through 128, 129 

Ornamental roof flange for flag pole 475 

Orthographic projection, Architectural definition 14 

Outside miter, Architectural definition 15 



PAGE 

Oval flaring object or hip bath 318 

or egg-shaped flaring pan 263 

P 

Pan, Automobile dust 420 

Elliptical flaring 260 

Elliptical or frustum of elliptical cone 317 

Pour-sided and flaring 104 

or flaring article, base oblong, top square 177 

Oval or egg-shaped flaring 263 

Eectangular flaring, rounded corners, ends flaring more 

than sides 316 

Panel, Architectural definition : 13 

molding, Architectural term 10 

patterns for face miter at right angles 106 

on face of racking bi'acket 196 

triangular in shape, Moldings bounding 108 

Paper, Drawing and tracing, description 2S, 29 

Parabola, Geometrical definition 7 

Parallel lines, Geometrical definition 2 

forms or miter cutting 96, 239 

Pattern cutting, Its application to sheet metal work 1 

Exemplification of the principles of 71, 95 

Sheet metal, Geometrical definition 1 

of frustum derived from wooden model 82 

derived from drawing 82 

of irregular shaped article by triangulation 90 

Pedestal, Architectural definition 11 

Equilateral triangle Ill 

Hexagon 114 

Octagonal 115 

Square 112 

Six-sided 114 

Pediment, Architectural definition 11 

Broken, Architectural definition 12 

Modified profile of horizontal return at top 202 

on a wash 164 

with modified profile and normal profile in horizontal re- 
turn 201 

or raking molding with normal profile and modified profile 

in return 200 

Baking bracket for curved 356 

Segmental broken, To obtain modified profiles in returns at 

foot and top 209 

Pencils, Lead, for drawing, description 25 

Pens, Puling, for drawing, description 26 

Pentagon, 'Geometrical definition 4 

regular, To draw upon given side 46 

To inscribe within given circle 43 

vase 112 

Pentagonal prism, Geometrical definition 8 

Perimeter, Geometrical definition 4 

Perpendicular at given point in straight line, To erect by 

compasses and straight edge 36 

lines, Geometrical definition 2 

near the end of a given straight line to erect by compasses 

and straight edge 36 

To erect to arc of circle without recourse to center 41 

Perspective projection, Architectural definition 14 

Pilaster, Architectural definition 1 

Gable molding mitered against molded 161 

Pillar, Architectural definition 11 

Pinnacle, Architectural definition 12 



530 



Index. 



PAGE 

Pinnacle, gable moldings and roof pieces, Octagonal 174 

moldings and roof pieces of square 173 

Octagon, with different pitches, Miter between moldings, 

of adjacent gables of 208 

with eight sides, Conical spire mitering on 282 

with four sides, Conical spire mitering on 280 

with rectangular shaft, Miter between moldings of adjacent 

gables of different pitches 207 

Pins, Drawing, or thumb tacks for drawing, description.... 27 

Pipe against roof of one inclination, Flange to fit around. . 129 
arms of same diameter as main pipe leaving at same angle, 

bifurcated 137 

at forty-five degrees, Two collars tapping 502 

circle to ellipse offset transition piece 452 

cut off obliquely 125 

elbows, see Elbows. 

intersecting cylinder obliquely, Rectangular 146 

four-piece elbow through one of the miters 159 

taper-joint axis not at right angles to axis or side of 

taper -joint 298 

axes parallel but to one side of each other 294 

axis crossing at right angles 272 

axis not at right angles 96 

Oblique offset in rectangular 210, 214 

offsetting blower, Connection for rectangular 507 

offsetting round to elliptical offset for 396 

over ridge of roof, Flange to fit around 128 

passing through apex of pyramid or hipped roof, Flange 

to fit against 140 

passing centrally through a pyramid or hipped roof, Flange 

fitting against 145 

Rectangle to round seven-piece elbow 405 

Rectangular, Designing elbow for 469 

intersecting a cylinder obliquely 146 

Oblique offsetting blower connection 507 

Transforming elbow for 473 

Reducer, Collar tapping main at an angle in two directions 501 

Double Offset Y-Branch for 466 

Five-piece elbow for 133 

flange to fit at top and bottom of flange resting in in- 
clined roof 363 

Four-piece elbow for 132 

Four-piece tapering elbow for 440 

intersecting elliptical pipe or smaller diameter at other 

than right angles, two cases 155 

intersecting straight back reducing joint at top 323 

joint between pipes of different diameters at other than 

a right angle, axis of smaller pipe at one side of larger 158 
joint between pipes of different diameters at right angles, 

axis of smaller pipe passing to side of larger 156 

joint between two diameters at other than right angle. . 151 

joint with straight back 311 

Junction of large pipe with elbows of two smaller ones. 384 

Oblique offsetting Y-branch for 510 

of unequal diameters, Collar to join at an angle 359 

on taper joints 292, 302 

on flag pole on inclined roof, Ornamental flange for. . . . 475 
passing through inclined roof, flange having equal projec- 
tion on all sides, Flange to fit 368 

round, carried around a semicircle by cross joints 134 

Tapering collar for axes crossing at right angles 270 

Tapering collar for axes crossing at other than right 

angle 290 



PAGE 

Pipe reducer, Tapering two-piece elbow for 284 

Three-piece elbow for 131 

T-joint between, at right angles of different diameters.. 153 

to fit against roof of one inclination 125 

to fit over ridge of roof 127 

to elliptical offsetting 396 

to elliptical two-piece elbow 287 

to oblong, semicircular ends, Offset for 393 

to octagon, transition piece 455 

- to rectangle, Seven-piece elbow for 405 

Two collars tapping main at forty-five degrees 505 

Two-piece elbow for 130 

with straight back for reducer joint 311 

Square, Compound curved elbows 478 

Describing twist or compound curve 140 

Designing elbows for 469 

Elbows for 494 

intersecting a cylinder obliquely 146 

oblique offsetting, blower connection 507 

Oblique offsetting elbows for 494 

to square, straight back transition 446 

Transforming elbow for 473 

twisted offset elbows for 497 

Twist or compound curve 140 

Tapering five-piece elbow for 287 

intersected at lower end by straight pipe of larger diam- 
eter, crossing at right angles 270 

intersecting one of larger diameter obliquely 302 

three-piece elbow for 2 

two-piece elbow for 284 

taper- joint for small, at one side of center line 334 

for straight back 341 

Tapping a straight back reducing joint off its top 323 

T-joint between pipes of same diameter 139 

Three-pieced elbow for round to elliptical 403 

three-pronged fork for 391 

to fit against a roof of one inclination 125 

to fit against a roof of one inclination, Elliptical 125 

to fit over ridge of roof 127 

True angle in compound offsetting elbows 147 

with circular ends, Round to oblong 393 

with semicircular ends, Taper-joint for oblong 258 

with semicircular ends, Round to oblong 482 

Y-braneh for 387 

Pipes of different diameters axis of smaller pipe passing to 

one side of the larger, T-joint between 156 

axis of smaller pipe placed to one side of larger one, 

Joint at other than right angles between 158 

intersecting at other than right angles, joint between. 154 

T-joint between 153 

of same diameter at other than right angles, Joint between. 151 

of unequal diameters, Collar joining at an angle 359 

Pitcher, Lip of sheet metal 272 

Planceer, Architectural definition 13 

Plain tapering keystone, Face and side of 104 

Plane, Geometrical definition 3 

figure, Geometrical definition 3 

Point, Geometrical definition . 1 

given, To draw tangent to circle or arc circle without re- 
course to center 41 

Pointed arch, Architectural term 11 

Points not in a straight line, To draw a circle through three. 41 

Polygon, Geometrical definition 3 



Index. 



531 



PAGE 

Polygon, finial, Irregular 185 

regular, General rule for miseubing within a given eircler. 44 
of eleven sides (undeeagon), To inscribe within given 

circle by rule 45 

rule for drawing when length of side is given 48 

Polygons, regular, Constructing by tractor 57 

Construction of by compasses and straight-edge 43 

To construct by T-Square and triangles or set-squares.. 49 

Polyhedron, Geometrical definition 9 

Portable-forge hood 371 

Prism, Geometrical definition 7 

Profile, Normal, Architectural definition 14 

of horizontal molding to miter with it at octagon angle 
and patterns for both arms, Prom the profile of 

given inclined molding to establish 206 

return at foot of gable to miter at right angles with 
inclined molding of normal profile and miter patterns 

of both, To obtain 200 

return at top of broken pediment to miter with inclined 

molding and patterns of both, To obtain 202 

of inclined molding to miter at right angles with horizontal 

return and miter patterns of both, To obtain 201 

of returns at top and foot of segmental broken pediment. . 204 

Eaked, Architectural definition 14 

to obtain the profile of inclined molding to miter with it at 

octagonal angle and patterns for both arms 205 

True, at line of intersection of curved or inclined molding 

against oblique flat surface 104 

Profiles in seven-sided figure, Changing 185 

Miter between two moldings of different 102 

Projection drawing, Architectural definition 14 

or linear drawing 30, 35 

Properties of circles, Geometry of 4 

Protractor for drawing regular polygons 57 

Protractors for drawing, description 24 

Pumps, Bosses for sheet metal hand 448 

Pyramid, Diamond shape, Envelope in frustum 178 

Envelope of hexagonal 241 

envelope of frustum of octagonal 243 

envelope of frustum of square 242 

Envelope of square 241 

Envelope of triangular 240 

Frustum of, Geometrical definition 9 

Geometrical definition 8 

having alternate long and short sides, Envelope of frustum 

of octagonal 244 

Square to fit against sides of elliptical pipe which passes 

through its center 145 

Pyramidal flange to fit sides of pipe passing through its apex 144 

Q 

Quadrant base and round top, Transitional object 354 

Geometrical definition 5 

Quadrangular prism, Geometrical definition 8 

Quadrilateral figure, Geometrical definition 3 

R 

Eadial line system of pattern cutting 7 

system of triangulation for elliptical cones 91, 93 

Radius. Geometrical definition 5 

Railing molding or broken pediment, modified profile of hori- 
zontal return at top 202 

Earn water cut-off 137 



PAGE 

Raised oblong cover with semicircular ends 258 

panel on face of raking bracket 196 

Rake miter, Architectural definition 15 

molding, Architectural definition 13 

Raked molding, Architectural definition 14 

profile or stay, Architectural definition 14 

Raking bracket 193, 356 

Raised panel or face of 196 

molding or pediment with modified profile and normal pro- 
file in horizontal return 201 

or pediment with normal profile and modified profile in 

return 200 

soffit bracket 508 

Range hood with reduced sides 512 

Rectangle base and round top transitional object, top not 

central over base 331 

and square top flaring article 177 

Geometrical definition 4 

pipe, Compound curved elbow for 478 

describing twist or compound curve 140 

Oblique offsetting elbows for 494 

Oblique twisted offsetting elbows for 497 

To draw an ellipse within, by intersecting lines 63 

to round straight back transition piece 446 

Rectangular flaring article 104 

Reduced return miters 182 

Reducer-joint for pipe 341 

small at one side of center line 339 

with straight back 311 

Requirements for cutting miters 76 

Return miter at other than a right angle, as in cornice at 

corner of building 98 

Returns at top and foot of segmental broken pediment 204 

Reversing joint for oblong pipe, semicircular ends 350 

Revolving blade of spiral conveyor 486 

Rhomboid, Geometrical definition 4 

Rhombus lozenge, Geometrical definition 4 

Ridge molding, Architectural definition 13 

Right angle, Geometrical definition 2 

Right-angle elbows in round pipe, two-, three-, four- and five- 
piece 1, 133 

Triangle, Geometrical definition 3 

Right cone, base oblique to its axis, Envelope of 265 

cut by plane parallel to its axis, Envelope of 278 

Envelope 249 

Envelope of frustum of 250 

contained between planes oblique to its axis, Envelope of 

frustum 274 

Geometrical definition 8 

upper plane oblique to its axis, Envelope of 265 

cylinder, Geographical definition 8 

pyramid, Geometrical definition 8 

Roof collar, round top, square base, to fit around pipe passing 

through inclined roof 333 

flange, Conical, to fit around pipe against inclined roof. . 266 
to fit around pipe against roof of double inclination. . 269 
Flaring, round bottom, top to fit pipe passing through 

inclined roof 363 

Ornamental flag pole 475 

to fit against sides of elliptical pipe passing centrally 

through a pyramid or hipped roof 145 

to fit against sides of pipe passing through apex of 
pyramid or hipped roof 144 



532 



Index. 



PAGE 

Roof flange, to fit around pipe against inclined roof 129 

to fit around pipe over ridge of roof " 128 

to lit pipe passing through inclined roof, flange to have 

equal projection from small sides 308 

Octagon, shaft mitering on ridge and hips of 127 

shaft to fit on ridge 126 

pieces and moldings in gables of square pinnacle 173 

Pipe to fit against inclined 125 

Pipe to fit on ridge 127 

Rubber or eraser for drawing, description 28 

Rule for cutting miters 77 

for inscribing any regular polygon within circle 44 

Rules, Pattern cutting 71 95 

Ruling pens for drawing, description 25 

S 

Scalene cone, Envelope 306 

Envelope frustum of 311 

intersected obliquely by a cylinder their axes not in the 

same plane, patterns of frustum of 323 

Scale scoop, having both ends alike, pattern 278 

one end of which is funnel shaped, patterns 279 

drawing, Architectural definition 14 

rules for drawing, description „ 27 

Geometrical definition 8 

system of triangulation for 94 

Triangle, Geometrical definition 3 

Scoop, Scale, both ends alike 278 

one end funnel shaped 279 

Scroll for bracket or modillion, To draw to a specified width. 69 

Scuttle, Funnel coal 399 

Secant, Geometrical definition 7 

Section, Conic, Geometrical definition 9 

drawing, Architectural definition 14 

of cone, Drawing to obtain an ellipse 62 

Sector, Geometrical definition 5 

Segment, Geometrical definition 5 

of a circle, Height and chord given, to find center from 

which arcs may be struck 40 

Segmental arch, Architectural term 11 

broken pediment, To obtain profile and patterns of returns 

at top and foot 204 

Semicircle, Geometrical definition 5 

Semicircle, Protractor for drawing, description 24 

Separator, Spiral strips for dust 432 

Spiral strip for dust, short rule 496 

Set squares or triangle, description 21 

Sets of centers, To draw approximate ellipses with compasses, 

using two or three 64 

Seven-piece square to round elbow 405 

Seven-sided vase 115 

Shaft, Architectural definition 11 

Miter between moldings of adjacent gables of different 

pitches upon pinnacle with rectangular 207 

Octagon, alternate long and short sides 189 

Sheet-metal pattern cutting, Geometrical definition 1 

Ship ventilator, round base, elliptical mouth 381 

Side, Pace of plain tapering keystone 104 

Geometrical definition 3 

Simple gable miters 110 

Sine, Geometrical definition 6 

Single curved surface, Geometrical definition 3 

Sink, Architectural definition 13 



PAGE 

Skylight bar, Top and bottom of common 138 

Top and bottom of hip bar 219 

Top of jack bar in 221 

Slabs for India ink for drawing, description 27 

Slant heights, Geometrical definition 8- 

Soap maker 's float 320 

Soffit, Architectural definition 13 

bracket, Raking 508 

of arch in circular wall, level at top and jambs of opening 

splayed on inside 416 

of semicircular arch in circular wall, level at top and jambs 

of opening at right angles to walls, two cases 408 

Solid, Geometrical definition 7 

of ordinary elliptical flaring article obtained by combin- 
ing parts of cones S4, 85 

Solids, Geometrical terms and definitions of 7 

Intersections of, Geometrical definition 9 

Spacers, Steel spring, description 22 

Sphere, Octagon shaft fitting against 122 

or globe, Geometrical definition 9 

Square shaft fitting against 122 

Spiral by spool and thread, To draw 69 

conveyor or loading spout 426 

Revolving blade for 486 

from centers with compasses, To draw 68 

strips in dust separator 432 

strip in dust separator, short rule 496 

Spire, Dormer molding mitering against octagonal side of... 165 

mitering upon eight gables. Conical 282 

upon eight gables, Octagon 245 

upon four gables, Conical 280 

upon four gables, Octagon 246 

upon four gables, Square 244 

upon roof at junction of ridge and hips, Octagon 248 

Splayed arch in circular wall, larger opening being on inside 

of wall 414 

elliptical arch in circular wall, larger opening on outside.. 412 

Spool and thread, To draw spiral with 69 

Spout bosses for sheet metal hand pumps 448 

Can breast 302 

intersecting cone hopper 492 

Loading, or spiral conveyor 426 

Watering can 290 

Springing lines, Architectural term 11 

Square about a given circle, To draw 57 

Geometrical definition 4 

face miter produced where square return miter was intended 78 

pinnacle, Moldings and roof pieces in 173 

pyramid, Envelope 241 

Envelope of frustum of 242 

to fit against sides of elliptical pipe passing through its 

center 145 

Square return miter, comparing short method of cutting square 

miter and method inscribed by rule 77 

Method of cutting 76 

or miter at right angles, as in corner cornice 98 

shaft to fit against sphere 122 

curved profile mitering over peak of gable coping having 

a double wash 169 

spire mitering upon four gables 244 

T, description 20 

To inscribe within given circle 43, 52, 58 

upon a given side, To draw 53, 57 



Indt 



x. 



533 



PAGE 
Starter or furnace boot, round to oblong, semicircular ends, 

one straight back, other with offset 34S 

round to rectangule double offsetting 443 

semicircular ends, openings central, Bound to oblong. 346 
offsetting furnaee boot, round to oblong, semicircular 

ends 393 

Stay, Architectural definition °. . . . 13 

Normal, Architectural definition 14 

Baked, Architectural definition 14 

Steel square, description 20 

for drawing circles 41 

testing for accuracy 21 

Stile, Architectural term 10 

Stilted- arch, Architectural definition 11 

Stoker, Automatic, or spiral conveyor, Revolving blade 4S6 

Stop block, Architectural definition 12 

Straight-edges, description 19 

Straight line, Geometrical definition 1 

sided figures, Geometrical terms and definitions 3 

Striking arc of circle by triangular guide, chord and height 

given 41 

Strip in dust separator, Spiral 432 

Spiral, in dust separator, short rule 496 

Surface, Butt miter against irregular or molded 103 

Double curved, Geometrical definition 3 

Geometrical definition 3 

Systems of pattern cutting 71, 95 

T 

Tables for drafting, description 17 

Tacks or drawing pins for drawing, description 27 

Tangent of an arc, Geometrical definition 6 

To draw, to circle or arc of circle at a given point without 

recourse to center 41 

Tank, Hood of oil 374 

Tapering article, square base, octagonal top 188 

with equal flare corresponding to frustum of cone, base 
an approximate ellipse struck from centers, upper plane 

of frustum oblique to axis 276 

elbow in five pieces, Begular 287 

five-piece elbow 287 

forms or "flaring work, Begular 79 

four-piece elbow 440 

joint intersected at large end by straight pipe, axes cross- 
ing at right angles 270 

intersected at large end by straight pipe at other than 

right angles 290 

intersected at small end by straight pipe axes crossing 

at right angles 275 

plain keystone, Face and side of 104 

three-piece elbow 285 

middle piece tapering 355 

two-piece elbow 284 

Taper-joint for round pipe, small end at side of center line. . 337 

iu oblong pipe with semicircular ends 258 

intersected at large end by straight pipe, axis of taper 

passing to one side of straight pipe 291 

obliquely intersecting another of larger diameter 302 

with round collar branching off at right angles to side 

of taper-joint 296 

with round collar branching off axis, collar not at right 

angles to axis nor side of taper-joint 298 

with round collar branching off, axis crossing at right 
angles 292 



PAGE 

Taper-joint with straight back 311 

with straight back for round pipe 341 

with tapering collar branching off, axis crossing at right 

angles 300 

Ten-sided newel post 117 

Testing drawing boards, Description of methods 19 

steel squares 21 

triangles or set squares 22 

Tetrahedron, Geometrical definition 9 

Thread and spool, to draw a spiral 69 

Thirteen-sided regular polygon, length of sides given, To 

construct by rule 48 

Three-piece elbow 131 

in tapering pipe 285 

middle piece being a gore 187 

middle piece tapering 355 

to join round with elliptical pipe 403 

Three-prong fork in two pieces 425 

with tapering branches 381 

Three sets of centers, to draw approximate ellipse with com- 
passes using 65 

Three-sided pedestal '. Ill 

Thumb tacks for drawing, description 27 

T- Joint between pipes of different diameters 153 

between pipes of different diameters, axis of smaller pipe 

passing to one side of the larger 156 

between pipes of same diameter, patterns 139 

smaller pipe tapering and intersected at large end by 

larger straight pipe, axes crossing at right angles 270 

Tombstone, Octagon, alternate long and short sides 1S9 

Tools and materials for drawing, Description 17 

Top elliptical, base rectangular 330 

for square to round chimney 327 

oblong, with semicircular ends and rectangular bottom.... 310 

octagonal, base square, tapering article 186 

rectangular, base round, base larger than top 335 

round, base elliptical, one side vertical, transitional object. 345 

and base elliptical, top and base central 343 

base quadrant, transitional object 354 

base rectangular, top not central over base 331 

base rectangular, top wider than narrow side of rect- 
angle, transitional object 328 

base square and inclined transitional object 333 

bottom elliptical, top and bottom not parallel 356 

bottom oblcng, semicircular ends, one with straight 

back ami another with offset 348 

bottom oblong, semicircular ends, top and bottom cen- 
tral 346 

square, bottom oblong, flaring article 177 

Tower, Dormer molding mitering against octagonal sides. . . . 165 

Tracing paper and cloth for drawing, description 29 

Trammels or beam eompasses, description 23 

Transformer, Collar intersecting 4S8 

Transforming square elbow 473 

Transition from rectangular base to elliptical top, Article 

forming 330' 

from round horizontal base to round top placed vertically. 376 



piece . 



503 

Circle to ellipse offset 452 

from square base to octagonal top 1S8 

gore in molded article forming from a square to an 

octagon 192 

gore octagon to square, as at end of a chamfer 191 

joining two pipes of unequal diameter at an angle 359 



534 



Index. 



PAGE 

Transition piece, octagon to round 455 

rectangular at one end, round at the other, round end not 

parallel to rectangular end 337 

rectangular base, round top, both central 327 

rectangular base, round top, diameter of circle greater 

than narrow side of rectangle 328 

rectangular base to round top, top not central over base. 331 

rectangular top, round base, base larger than top 335 

round to quadrant 35-1 

round top, elliptical bottom, both central 343 

round top, elliptical bottom, one side vertical 345 

round top, oblong bottom, ends concentric 346 

round top, bottom oblong, semicircular ends, two cases. . 312 

round top, oblong bottom, two cases 348 

round top, rectangular base, inclined 333 

square to octagon 455 

square to round elbow 431 

square to round, circle wider than square 328 

square to round, top and bottom central 327 

square to round, top not central over base 331 

square to round, straight back 446 

top square, base round, base larger than top 335 

Trapezium, Geometrical definition 4 

Trapezoid, Geometrical definition 4 

Tray, heart-shaped flaring 261 

Triangle, acute-angled. Geometrical definition 3 

Equilateral, Geometrical definition 3 

To construct upon given side 45, 53 

To draw about given circle 55 

To draw upon a given side 53 

To draw without a given circle 57 

To inscribe within a given circle 43, 51 

Geometrical definition 3 

Isosceles, Geometrical definition 3 

length of three sides given. To construct 45 

Obtuse angled, Geometrical definition 3 

or set square, description 21 

Proper method of using a 45-degree to divide a circle 51 

Proper method of using a 30X60 degree triangle for divid- 
ing a circle 52 

Eight-angled, Geometrical definition 3 

Scalene, Geometrical definition 3 

Triangular panel, Face miters of moldings 108 

prism, Geometrical definition 7 

pyramid envelope 240 

scale rule for drawing, description 25 

Triangulation by radial line systems for elliptical cones. .. 91, 93 

obtaining pattern by 90 

or irregular shaped forms 86, 95 

system for scalene cones 94 

Trisecting an angle 39 

True augle at hips of irregular square box 477 

in compound offsetting elbows of round piping 143 

Truncated cone, Geometrical definition 8 

Truncating cone obliquely 81 

Truss or head block, Architectural definition 12 

T-Square, description 20 

Twelve-sided elliptical vase 1S3 

urn 118 

Twisted elbow for square pipe 478 

oblique offsetting elbows in square pipe 497 

or compound curbed elbow in square pipe 140 

Two-piece circle to ellipse offset transition elbow 452 

elbow 130 



PAGE 

Two-piece elbow in tapering pipe 284 

one end round and the other elliptical, Eight angle 387 

in an elliptical pipe, two cases 130 

three-prong fork 425 

Two-pronged double offset fork 466 

Two-prong fork for pipe 389 

Two sets of centers, To draw approximate ellipse with com- 
passes, using 64 

Two-way Y, Developing patterns for unusual 429 

Tub, Flaring end of oblong 178 

tapering sides, semicircular, head flaring more than sides 314 

Hip bath 318 

Portable bath 366 

U 
Undecagon or eleven-sided figure, To inscribe within given 

circle by general rule 45 

Eegular, To draw upon given side , 48 

To inscribe within given circle 44 

Urn, Dodecagon 118 

Five-sided true polygonal 112 

Seven-sided true polygonal 115 

Twelve-sided elliptical 183 

V 

Vase, Heptagon 115 

Pentagon 112 

Twelve-sided elliptical 183 

Ventilation pipe, Oblique twisted offsetting elbows 497 

work. Compound curved elbow in square 140 

Ventilator, Ship, round base, elliptical mouth 381 

Versed line, Geometrical definition 7 

Vertex, Geometrical definition 3 

Vertical lines, Geometrical definition 2 

Volute, Architectural definition 13 

Construction for a capital 142 

Geometrical term 67 

To draw Ionic 67 

To draw simple 67 

Voussoirs, Architectural term 11 

W 

Wash, including a return, Inclined molding mitering upon. . 166 

pediment 164 

Wheel, Special design, fender for automobiles 462 

Window cap, Simple curved moldings in 255 

Wooden cone, plan and elevation from which to construct .... 79 

model, Method of deriving pattern of frustum from 82 

Words, Architectural and Geometrical, Alphabetical list. . 15, 16 

Working drawing, Architectural definition 14 

Y 
Y-branch, arms of same diameters as main pipe, leaving it 

at same angle 137 

Double offsetting 466 

for round pipe, Oblique offsetting 510 

Unusual two-way 429 

Y-branches, forty-five degree 505 

Y, consisting of tapering pipes joining larger pipe at an angle 389 



Zones, To construct ball in any number of pieces of shape. . 251 



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